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Solving problems which involve forces, friction, and Newton's Laws: A step-by-step guide

This step-by-step guide is meant to show you how to approach problems where you have to deal with moving objects subject to friction and other forces, and you need to apply Newton's Laws. We will go through many problems, so you can have a clear idea of the process involved in solving them.

The problems we will examine include objects that

  • are pushed/pulled horizontally with an angle
  • move up or down an incline
  • hang from ropes attached to the ceiling
  • hang from ropes that run over pulleys
  • move connected by a string
  • are pushed in contact with each other (Coming soon!)
  • Box pulled at an angle over a horizontal surface
  • Block pushed over the floor with a downward and forward force
  • Object moving at constant velocity over a horizontal surface
  • Block pushed up a frictionless ramp
  • Mass pulled up an incline with friction
  • A mass hanging from two ropes
  • Two hanging objects connected by a rope
  • Two masses on a pulley
  • Two blocks connected by a string are pulled horizontally

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High School Physics : Calculating Force

Study concepts, example questions & explanations for high school physics, all high school physics resources, example questions, example question #1 : net force.

problem solving in force physics

Plug these into the equation to solve for acceleration.

problem solving in force physics

Example Question #2 : Calculating Force

problem solving in force physics

Plug in the values given to us and solve for the force.

problem solving in force physics

Example Question #3 : Calculating Force

problem solving in force physics

Plug in the given values to solve for the mass.

problem solving in force physics

Example Question #4 : Calculating Force

problem solving in force physics

(Assume the only two forces acting on the object are friction and Derek).

problem solving in force physics

Plug in the information we've been given so far to find the force of friction.

problem solving in force physics

Friction will be negative because it acts in the direction opposite to the force of Derek.

Example Question #5 : Calculating Force

problem solving in force physics

Newton's third law states that when one object exerts a force on a second object, the second object exerts a force equal in size, but opposite in direction to the first. That means that the force of the hammer on the nail and the nail on the hammer will be equal in size, but opposite in direction.

problem solving in force physics

Example Question #6 : Calculating Force

problem solving in force physics

We can find the net force by adding the individual force together.

problem solving in force physics

Example Question #7 : Calculating Force

problem solving in force physics

If the object has a constant velocity, that means that the net acceleration must be zero.

problem solving in force physics

In conjunction with Newton's second law, we can see that the net force is also zero. If there is no net acceleration, then there is no net force.

problem solving in force physics

Since Franklin is lifting the weight vertically, that means there will be two force acting upon the weight: his lifting force and gravity. The net force will be equal to the sum of the forces acting on the weight.

problem solving in force physics

We know the mass of the weight and we know the acceleration, so we can solve for the lifting force.

problem solving in force physics

Example Question #1 : Calculating Force

problem solving in force physics

We are given the mass, but we will need to calculate the acceleration to use in the formula.

problem solving in force physics

Plug in our given values and solve for acceleration.

problem solving in force physics

Now we know both the acceleration and the mass, allowing us to solve for the force.

problem solving in force physics

Example Question #9 : Calculating Force

problem solving in force physics

We can calculate the gravitational force using the mass.

problem solving in force physics

Example Question #10 : Calculating Force

problem solving in force physics

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Dynamics: Force and Newton’s Laws of Motion

Problem-solving strategies, learning objective.

By the end of this section, you will be able to:

  • Understand and apply a problem-solving procedure to solve problems using Newton’s laws of motion.

Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy for Newton’s Laws of Motion

Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation . Such a sketch is shown in Figure 1(a). Then, as in Figure 1(b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists).

Figure 1. (a) A sketch of Tarzan hanging from a vine. (b) Arrows are used to represent all forces. T is the tension in the vine above Tarzan, F T is the force he exerts on the vine, and w is his weight. All other forces, such as the nudge of a breeze, are assumed negligible. (c) Suppose we are given the ape man’s mass and asked to find the tension in the vine. We then define the system of interest as shown and draw a free-body diagram. F T is no longer shown, because it is not a force acting on the system of interest; rather, F T  acts on the outside world. (d) Showing only the arrows, the head-to-tail method of addition is used. It is apparent that T = –w , if Tarzan is stationary.

Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest . This decision is a crucial step, since Newton’s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 1(c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well.

A diagram showing the system of interest and all of the external forces is called a free-body diagram . Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure 1(c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem . This is done in Figure 1(d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known.

Applying Newton’s Second Law

image

F net x  = ma ,

F net y = 0.

You will need this information in order to determine unknown forces acting in a system.

Step 4. As always, check the solution to see whether it is reasonable . In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake.

Section Summary

To solve problems involving Newton’s laws of motion, follow the procedure described:

  • Draw a sketch of the problem.
  • Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.
  • Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x-direction) then  F net x  = 0 . If the object does accelerate in that direction,  F net x  = ma .
  • Check your answer. Is the answer reasonable? Are the units correct?

Problems & Exercises

1. A 5.00 × 10 5 -kg rocket is accelerating straight up. Its engines produce 1.250 × 10 7  of thrust, and air resistance is 4.50 × 10 6 N. What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

2. The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is 1.80 m/s 2 , what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. For this situation, draw a free-body diagram and write the net force equation.

3. Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

4. When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

5. A freight train consists of two 8.00 × 10 4   engines and 45 cars with average masses of 5.50 × 10 4 kg . (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00 × 10 -2  if the force of friction is 7.50 × 10 5 , assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

6. Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) An 1800-kg tractor exerts a force of 1.75 × 10 5  backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is 0.150 m/s 2 , what is the mass of the airplane? (b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. (c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.

7. A 1100-kg car pulls a boat on a trailer. (a) What total force resists the motion of the car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an acceleration of 0.550 m/s 2 ? The mass of the boat plus trailer is 700 kg. (b) What is the force in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?

8. (a) Find the magnitudes of the forces F 1 and F 2  that add to give the total force F tot  shown in Figure 4. This may be done either graphically or by using trigonometry. (b) Show graphically that the same total force is obtained independent of the order of addition of   F 1 and F 2 . (c) Find the direction and magnitude of some other pair of vectors that add to give F tot . Draw these to scale on the same drawing used in part (b) or a similar picture.

A right triangle is shown made up of three vectors. The first vector, F sub one, is along the triangle’s base toward the right; the second vector, F sub two, is along the perpendicular side pointing upward; and the third vector, F sub tot, is along the hypotenuse pointing up the incline. The magnitude of F sub tot is twenty newtons. In a free-body diagram, F sub one is shown by an arrow pointing right and F sub two is shown by an arrow acting vertically upward.

9. Two children pull a third child on a snow saucer sled exerting forces  F 1 and F 2 as shown from above in Figure 4 . Find the acceleration of the 49.00-kg sled and child system. Note that the direction of the frictional force is unspecified; it will be in the opposite direction of the sum of  F 1 and F 2 .

An overhead view of a child sitting on a snow saucer sled. Two forces, F sub one equal to ten newtons and F sub two equal to eight newtons, are acting toward the right. F sub one makes an angle of forty-five degrees from the x axis and F sub two makes an angle of thirty degrees from the x axis in a clockwise direction. A friction force f is equal to seven point five newtons, shown by a vector pointing in negative x direction. In the free-body diagram, F sub one and F sub two are shown by arrows toward the right, making a forty-five degree angle above the horizontal and a thirty-degree angle below the horizontal respectively. The friction force f is shown by an arrow along the negative x axis.

10. Suppose your car was mired deeply in the mud and you wanted to use the method illustrated in Figure 6 to pull it out. (a) What force would you have to exert perpendicular to the center of the rope to produce a force of 12,000 N on the car if the angle is 2.00°? In this part, explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. (b) Real ropes stretch under such forces. What force would be exerted on the car if the angle increases to 7.00° and you still apply the force found in part (a) to its center?

Figure of car stuck in the mud and a rope connected to a tree trunk in an attempt to pull out the car.

11. What force is exerted on the tooth in Figure 7 if the tension in the wire is 25.0 N? Note that the force applied to the tooth is smaller than the tension in the wire, but this is necessitated by practical considerations of how force can be applied in the mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for Newton’s laws of motion.

Cross-section of jaw with sixteen teeth is shown. Braces are along the outside of the teeth. Three forces are acting on the protruding tooth. The applied force, F sub app, is shown by an arrow vertically downward; a second force, T, is shown by an arrow making an angle of fifteen degrees below the positive x axis; and a third force, T, is shown by an arrow making an angle of fifteen degrees below the negative x axis.

Figure 7. Braces are used to apply forces to teeth to realign them. Shown in this figure are the tensions applied by the wire to the protruding tooth. The total force applied to the tooth by the wire, F app , points straight toward the back of the mouth.

12. Figure 9 shows Superhero and Trusty Sidekick hanging motionless from a rope. Superhero’s mass is 90.0 kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the rope is negligible. (a) Draw a free-body diagram of the situation showing all forces acting on Superhero, Trusty Sidekick, and the rope. (b) Find the tension in the rope above Superhero. (c) Find the tension in the rope between Superhero and Trusty Sidekick. Indicate on your free-body diagram the system of interest used to solve each part.

Two caped superheroes hang on a rope suspended vertically from a bar.

Figure 9. Superhero and Trusty Sidekick hang motionless on a rope as they try to figure out what to do next. Will the tension be the same everywhere in the rope?

13. A nurse pushes a cart by exerting a force on the handle at a downward angle 35.0º below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N. (a) Draw a free-body diagram for the system of interest. (b) What force must the nurse exert to move at a constant velocity?

14. Construct Your Own Problem Consider the tension in an elevator cable during the time the elevator starts from rest and accelerates its load upward to some cruising velocity. Taking the elevator and its load to be the system of interest, draw a free-body diagram. Then calculate the tension in the cable. Among the things to consider are the mass of the elevator and its load, the final velocity, and the time taken to reach that velocity.

15. Construct Your Own Problem Consider two people pushing a toboggan with four children on it up a snow-covered slope. Construct a problem in which you calculate the acceleration of the toboggan and its load. Include a free-body diagram of the appropriate system of interest as the basis for your analysis. Show vector forces and their components and explain the choice of coordinates. Among the things to be considered are the forces exerted by those pushing, the angle of the slope, and the masses of the toboggan and children.

16. Unreasonable Results (a) Repeat Exercise 7, but assume an acceleration of 1.20 m/s 2  is produced. (b) What is unreasonable about the result? (c) Which premise is unreasonable, and why is it unreasonable?

17. Unreasonable Results (a) What is the initial acceleration of a rocket that has a mass of 1.50 × 10 6  at takeoff, the engines of which produce a thrust of 2.00 × 10 6 ? Do not neglect gravity. (b) What is unreasonable about the result? (This result has been unintentionally achieved by several real rockets.) (c) Which premise is unreasonable, or which premises are inconsistent? (You may find it useful to compare this problem to the rocket problem earlier in this section.)

Selected Solutions to Problems & Exercises

1. Using the free-body diagram:

An object of mass m is shown. Three forces acting on it are tension T, shown by an arrow acting vertically upward, and friction f and gravity m g, shown by two arrows acting vertically downward.

  • [latex]{F}_{\text{net}}=T-f-mg=\text{ma}\\[/latex] ,

[latex]a=\frac{T-f-\text{mg}}{m}=\frac{1\text{.}\text{250}\times {\text{10}}^{7}\text{N}-4.50\times {\text{10}}^{\text{6}}N-\left(5.00\times {\text{10}}^{5}\text{kg}\right)\left(9.{\text{80 m/s}}^{2}\right)}{5.00\times {\text{10}}^{5}\text{kg}}=\text{6.20}{\text{m/s}}^{2}\\[/latex]

3. Use Newton’s laws of motion.

Two forces are acting on an object of mass m: F, shown by an arrow pointing upward, and its weight w, shown by an arrow pointing downward. Acceleration a is represented by a vector arrow pointing upward. The figure depicts the forces acting on a high jumper.

[latex]F=\left(\text{70.0 kg}\right)\left[\left(\text{39}\text{.}{\text{2 m/s}}^{2}\right)+\left(9\text{.}{\text{80 m/s}}^{2}\right)\right]\\[/latex] [latex]=3.\text{43}\times {\text{10}}^{3}\text{N}\\[/latex].  The force exerted by the high-jumper is actually down on the ground, but F is up from the ground and makes him jump.

  • This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of 10 3 N.

5. (a) 4.41 × 10 5 N (b) 1.50 × 10 5 N

7. (a) 910 N (b) 1.11 × 10 3

9. (a) a = 0.139 m/s, θ = 12.4º

11. Use Newton’s laws since we are looking for forces.

  • Draw a free-body diagram:

A horizontal dotted line with two vectors extending downward from the mid-point of the dotted line, both at angles of fifteen degrees. A third vector points straight downward from the intersection of the first two angles, bisecting them; it is perpendicular to the dotted line.

  • The tension is given as T = 25.0 N. Find F app . Using Newton’s laws gives:[latex]\sigma{F}_{y}=0\\[/latex], so that applied force is due to the y -components of the two tensions: F app = 2 T  sin  θ  = 2(25.0 N) sin(15º) = 12.9 N The x -components of the tension cancel. [latex]\sum{F}_{x}=0\\[/latex].
  • This seems reasonable, since the applied tensions should be greater than the force applied to the tooth.
  • College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License

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Video transcript

  • 8.1 Linear Momentum, Force, and Impulse
  • Introduction
  • 1.1 Physics: Definitions and Applications
  • 1.2 The Scientific Methods
  • 1.3 The Language of Physics: Physical Quantities and Units
  • Section Summary
  • Key Equations
  • Concept Items
  • Critical Thinking Items
  • Performance Task
  • Multiple Choice
  • Short Answer
  • Extended Response
  • 2.1 Relative Motion, Distance, and Displacement
  • 2.2 Speed and Velocity
  • 2.3 Position vs. Time Graphs
  • 2.4 Velocity vs. Time Graphs
  • 3.1 Acceleration
  • 3.2 Representing Acceleration with Equations and Graphs
  • 4.2 Newton's First Law of Motion: Inertia
  • 4.3 Newton's Second Law of Motion
  • 4.4 Newton's Third Law of Motion
  • 5.1 Vector Addition and Subtraction: Graphical Methods
  • 5.2 Vector Addition and Subtraction: Analytical Methods
  • 5.3 Projectile Motion
  • 5.4 Inclined Planes
  • 5.5 Simple Harmonic Motion
  • 6.1 Angle of Rotation and Angular Velocity
  • 6.2 Uniform Circular Motion
  • 6.3 Rotational Motion
  • 7.1 Kepler's Laws of Planetary Motion
  • 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity
  • 8.2 Conservation of Momentum
  • 8.3 Elastic and Inelastic Collisions
  • 9.1 Work, Power, and the Work–Energy Theorem
  • 9.2 Mechanical Energy and Conservation of Energy
  • 9.3 Simple Machines
  • 10.1 Postulates of Special Relativity
  • 10.2 Consequences of Special Relativity
  • 11.1 Temperature and Thermal Energy
  • 11.2 Heat, Specific Heat, and Heat Transfer
  • 11.3 Phase Change and Latent Heat
  • 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium
  • 12.2 First law of Thermodynamics: Thermal Energy and Work
  • 12.3 Second Law of Thermodynamics: Entropy
  • 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators
  • 13.1 Types of Waves
  • 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period
  • 13.3 Wave Interaction: Superposition and Interference
  • 14.1 Speed of Sound, Frequency, and Wavelength
  • 14.2 Sound Intensity and Sound Level
  • 14.3 Doppler Effect and Sonic Booms
  • 14.4 Sound Interference and Resonance
  • 15.1 The Electromagnetic Spectrum
  • 15.2 The Behavior of Electromagnetic Radiation
  • 16.1 Reflection
  • 16.2 Refraction
  • 16.3 Lenses
  • 17.1 Understanding Diffraction and Interference
  • 17.2 Applications of Diffraction, Interference, and Coherence
  • 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge
  • 18.2 Coulomb's law
  • 18.3 Electric Field
  • 18.4 Electric Potential
  • 18.5 Capacitors and Dielectrics
  • 19.1 Ohm's law
  • 19.2 Series Circuits
  • 19.3 Parallel Circuits
  • 19.4 Electric Power
  • 20.1 Magnetic Fields, Field Lines, and Force
  • 20.2 Motors, Generators, and Transformers
  • 20.3 Electromagnetic Induction
  • 21.1 Planck and Quantum Nature of Light
  • 21.2 Einstein and the Photoelectric Effect
  • 21.3 The Dual Nature of Light
  • 22.1 The Structure of the Atom
  • 22.2 Nuclear Forces and Radioactivity
  • 22.3 Half Life and Radiometric Dating
  • 22.4 Nuclear Fission and Fusion
  • 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation
  • 23.1 The Four Fundamental Forces
  • 23.2 Quarks
  • 23.3 The Unification of Forces
  • A | Reference Tables

Section Learning Objectives

By the end of this section, you will be able to do the following:

  • Describe momentum, what can change momentum, impulse, and the impulse-momentum theorem
  • Describe Newton’s second law in terms of momentum
  • Solve problems using the impulse-momentum theorem

Teacher Support

The learning objectives in this section will help your students master the following standards:

  • (C) calculate the mechanical energy of, power generated within, impulse applied to, and momentum of a physical system.

Section Key Terms

[BL] [OL] Review inertia and Newton’s laws of motion.

[AL] Start a discussion about movement and collision. Using the example of football players, point out that both the mass and the velocity of an object are important considerations in determining the impact of collisions. The direction as well as the magnitude of velocity is very important.

Momentum, Impulse, and the Impulse-Momentum Theorem

Linear momentum is the product of a system’s mass and its velocity . In equation form, linear momentum p is

You can see from the equation that momentum is directly proportional to the object’s mass ( m ) and velocity ( v ). Therefore, the greater an object’s mass or the greater its velocity, the greater its momentum. A large, fast-moving object has greater momentum than a smaller, slower object.

Momentum is a vector and has the same direction as velocity v . Since mass is a scalar , when velocity is in a negative direction (i.e., opposite the direction of motion), the momentum will also be in a negative direction; and when velocity is in a positive direction, momentum will likewise be in a positive direction. The SI unit for momentum is kg m/s.

Momentum is so important for understanding motion that it was called the quantity of motion by physicists such as Newton. Force influences momentum, and we can rearrange Newton’s second law of motion to show the relationship between force and momentum.

Recall our study of Newton’s second law of motion ( F net = m a ). Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. The change in momentum is the difference between the final and initial values of momentum.

In equation form, this law is

where F net is the net external force, Δ p Δ p is the change in momentum, and Δ t Δ t is the change in time.

We can solve for Δ p Δ p by rearranging the equation

F net Δ t F net Δ t is known as impulse and this equation is known as the impulse-momentum theorem . From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts. It is equal to the change in momentum. The effect of a force on an object depends on how long it acts, as well as the strength of the force. Impulse is a useful concept because it quantifies the effect of a force. A very large force acting for a short time can have a great effect on the momentum of an object, such as the force of a racket hitting a tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time.

[OL] [AL] Explain that a large, fast-moving object has greater momentum than a smaller, slower object. This quality is called momentum.

[BL] [OL] Review the equation of Newton’s second law of motion. Point out the two different equations for the law.

Newton’s Second Law in Terms of Momentum

When Newton’s second law is expressed in terms of momentum, it can be used for solving problems where mass varies, since Δ p = Δ ( m v ) Δ p = Δ ( m v ) . In the more traditional form of the law that you are used to working with, mass is assumed to be constant. In fact, this traditional form is a special case of the law, where mass is constant. F net = m a F net = m a is actually derived from the equation:

For the sake of understanding the relationship between Newton’s second law in its two forms, let’s recreate the derivation of F net = m a F net = m a from

by substituting the definitions of acceleration and momentum.

The change in momentum Δ p Δ p is given by

If the mass of the system is constant, then

By substituting m Δ v m Δ v for Δ p Δ p , Newton’s second law of motion becomes

for a constant mass.

we can substitute to get the familiar equation

when the mass of the system is constant.

[BL] [OL] [AL] Show the two different forms of Newton’s second law and how one can be derived from the other.

Tips For Success

We just showed how F net = m a F net = m a applies only when the mass of the system is constant. An example of when this formula would not apply would be a moving rocket that burns enough fuel to significantly change the mass of the rocket. In this case, you can use Newton’s second law expressed in terms of momentum to account for the changing mass without having to know anything about the interaction force by the fuel on the rocket.

Hand Movement and Impulse

In this activity you will experiment with different types of hand motions to gain an intuitive understanding of the relationship between force, time, and impulse.

  • one tub filled with water
  • Try catching a ball while giving with the ball, pulling your hands toward your body.
  • Next, try catching a ball while keeping your hands still.
  • Hit water in a tub with your full palm. Your full palm represents a swimmer doing a belly flop.
  • After the water has settled, hit the water again by diving your hand with your fingers first into the water. Your diving hand represents a swimmer doing a dive.
  • Explain what happens in each case and why.
  • a football player colliding with another, or a car moving at a constant velocity
  • a car moving at a constant velocity, or an object moving in the projectile motion
  • a car moving at a constant velocity, or a racket hitting a ball
  • a football player colliding with another, or a racket hitting a ball

[OL] [AL] Discuss the impact one feels when one falls or jumps. List the factors that affect this impact.

Links To Physics

Engineering: saving lives using the concept of impulse.

Cars during the past several decades have gotten much safer. Seat belts play a major role in automobile safety by preventing people from flying into the windshield in the event of a crash. Other safety features, such as airbags, are less visible or obvious, but are also effective at making auto crashes less deadly (see Figure 8.2 ). Many of these safety features make use of the concept of impulse from physics. Recall that impulse is the net force multiplied by the duration of time of the impact. This was expressed mathematically as Δ p = F net Δ t Δ p = F net Δ t .

Airbags allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant whether an airbag is deployed or not. But the force that brings the occupant to a stop will be much less if it acts over a larger time. By rearranging the equation for impulse to solve for force F net = Δ p Δ t , F net = Δ p Δ t , you can see how increasing Δ t Δ t while Δ p Δ p stays the same will decrease F net . This is another example of an inverse relationship. Similarly, a padded dashboard increases the time over which the force of impact acts, thereby reducing the force of impact.

Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that a car will crumple in a collision , especially in the event of a head-on collision. A longer collision time means the force on the occupants of the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident.

Grasp Check

You may have heard the advice to bend your knees when jumping. In this example, a friend dares you to jump off of a park bench onto the ground without bending your knees. You, of course, refuse. Explain to your friend why this would be a foolish thing. Show it using the impulse-momentum theorem.

  • Bending your knees increases the time of the impact, thus decreasing the force.
  • Bending your knees decreases the time of the impact, thus decreasing the force.
  • Bending your knees increases the time of the impact, thus increasing the force.
  • Bending your knees decreases the time of the impact, thus increasing the force.

Solving Problems Using the Impulse-Momentum Theorem

Talk about the different strategies to be used while solving problems. Make sure that students know the assumptions made in each equation regarding certain quantities being constant or some quantities being negligible.

Worked Example

Calculating momentum: a football player and a football.

(a) Calculate the momentum of a 110 kg football player running at 8 m/s. (b) Compare the player’s momentum with the momentum of a 0.410 kg football thrown hard at a speed of 25 m/s.

No information is given about the direction of the football player or the football, so we can calculate only the magnitude of the momentum, p . (A symbol in italics represents magnitude.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum:

To find the player’s momentum, substitute the known values for the player’s mass and speed into the equation.

To find the ball’s momentum, substitute the known values for the ball’s mass and speed into the equation.

The ratio of the player’s momentum to the ball’s momentum is

Although the ball has greater velocity, the player has a much greater mass. Therefore, the momentum of the player is about 86 times greater than the momentum of the football.

Calculating Force: Venus Williams’ Racquet

During the 2007 French Open, Venus Williams ( Figure 8.3 ) hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What was the average force exerted on the 0.057 kg tennis ball by Williams’ racquet? Assume that the ball’s speed just after impact was 58 m/s, the horizontal velocity before impact is negligible, and that the ball remained in contact with the racquet for 5 ms (milliseconds).

Recall that Newton’s second law stated in terms of momentum is

As noted above, when mass is constant, the change in momentum is given by

where v f is the final velocity and v i is the initial velocity. In this example, the velocity just after impact and the change in time are given, so after we solve for Δ p Δ p , we can use F net = Δ p Δ t F net = Δ p Δ t to find the force.

To determine the change in momentum, substitute the values for mass and the initial and final velocities into the equation above.

Now we can find the magnitude of the net external force using F net = Δ p Δ t F net = Δ p Δ t

This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact. This problem could also be solved by first finding the acceleration and then using F net = m a , but we would have had to do one more step. In this case, using momentum was a shortcut.

Practice Problems

  • 0.5 kg ⋅ m/s
  • 15 kg ⋅ m/s
  • 50 kg ⋅ m/s

A 155-g baseball is incoming at a velocity of 25 m/s. The batter hits the ball as shown in the image. The outgoing baseball has a velocity of 20 m/s at the angle shown.

What is the magnitudde of the impulse acting on the ball during the hit?

  • 2.68 kg⋅m/s.
  • 5.42 kg⋅m/s.
  • 6.05 kg⋅m/s.
  • 8.11 kg⋅m/s.

Check Your Understanding

What is linear momentum?

  • the sum of a system’s mass and its velocity
  • the ratio of a system’s mass to its velocity
  • the product of a system’s mass and its velocity
  • the product of a system’s moment of inertia and its velocity

If an object’s mass is constant, what is its momentum proportional to?

  • Its velocity
  • Its displacement
  • Its moment of inertia

What is the equation for Newton’s second law of motion, in terms of mass, velocity, and time, when the mass of the system is constant?

  • F net = Δ v Δ m Δ t F net = Δ v Δ m Δ t
  • F net = m Δ t Δ v F net = m Δ t Δ v
  • F net = m Δ v Δ t F net = m Δ v Δ t
  • F net = Δ m Δ v Δ t F net = Δ m Δ v Δ t

Give an example of a system whose mass is not constant.

  • A spinning top
  • A baseball flying through the air
  • A rocket launched from Earth
  • A block sliding on a frictionless inclined plane

Use the Check Your Understanding questions to assess whether students master the learning objectives of this section. If students are struggling with a specific objective, the assessment will help identify which objective is causing the problem and direct students to the relevant content.

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Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-physics . Changes were made to the original material, including updates to art, structure, and other content updates.

Access for free at https://openstax.org/books/physics/pages/1-introduction
  • Authors: Paul Peter Urone, Roger Hinrichs
  • Publisher/website: OpenStax
  • Book title: Physics
  • Publication date: Mar 26, 2020
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/physics/pages/1-introduction
  • Section URL: https://openstax.org/books/physics/pages/8-1-linear-momentum-force-and-impulse

© Jan 19, 2024 Texas Education Agency (TEA). The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

  • (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

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1.4: Solving Physics Problems

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Dimensional Analysis

Any physical quantity can be expressed as a product of a combination of the basic physical dimensions.

learning objectives

  • Calculate the conversion from one kind of dimension to another

The dimension of a physical quantity indicates how it relates to one of the seven basic quantities. These fundamental quantities are:

  • [A] Current
  • [K] Temperature
  • [mol] Amount of a Substance
  • [cd] Luminous Intensity

As you can see, the symbol is enclosed in a pair of square brackets. This is often used to represent the dimension of individual basic quantity. An example of the use of basic dimensions is speed, which has a dimension of 1 in length and -1 in time; \(\mathrm{\frac{[L]}{[T]}=[LT^{−1}]}\). Any physical quantity can be expressed as a product of a combination of the basic physical dimensions.

Dimensional analysis is the practice of checking relations between physical quantities by identifying their dimensions. The dimension of any physical quantity is the combination of the basic physical dimensions that compose it. Dimensional analysis is based on the fact that physical law must be independent of the units used to measure the physical variables. It can be used to check the plausibility of derived equations, computations and hypotheses.

Derived Dimensions

The dimensions of derived quantities may include few or all dimensions in individual basic quantities. In order to understand the technique to write dimensions of a derived quantity, we consider the case of force. Force is defined as:

\[\begin{align} \mathrm{F} &= \mathrm{m⋅a} \\ \mathrm{F} &= \mathrm{[M][a]} \end{align}\]

The dimension of acceleration, represented as [a], is itself a derived quantity being the ratio of velocity and time. In turn, velocity is also a derived quantity, being ratio of length and time.

\[\begin{align} \mathrm{F} &= \mathrm{[M][a]=[M][vT^{−1}]} \\ \mathrm{F} &= \mathrm{[M][LT^{−1}T^{−1}]=[MLT^{−2}]} \end{align}\]

Dimensional Conversion

In practice, one might need to convert from one kind of dimension to another. For common conversions, you might already know how to convert off the top of your head. But for less common ones, it is helpful to know how to find the conversion factor:

\[\mathrm{Q=n_1u_1=n_2u_2}\]

where n represents the amount per u dimensions. You can then use ratios to figure out the conversion:

\[\mathrm{n_2=\dfrac{u_2}{u_1}⋅n_1}\]

Trigonometry

Trigonometry is central to the use of free body diagrams, which help visually represent difficult physics problems.

  • Explain why trigonometry is useful in determining horizontal and vertical components of forces

Trigonometry and Solving Physics Problems

In physics, most problems are solved much more easily when a free body diagram is used. Free body diagrams use geometry and vectors to visually represent the problem. Trigonometry is also used in determining the horizontal and vertical components of forces and objects. Free body diagrams are very helpful in visually identifying which components are unknown and where the moments are applied. They can help analyze a problem, whether it is static or dynamic.

When people draw free body diagrams, often not everything is perfectly parallel and perpendicular. Sometimes people need to analyze the horizontal and vertical components of forces and object orientation. When the force or object is not acting parallel to the x or y axis, people can employ basic trigonometry to use the simplest components of the action to analyze it. Basically, everything should be considered in terms of x and y , which sometimes takes some manipulation.

Free Body Diagram : The rod is hinged from a wall and is held with the help of a string.

A rod ‘AB’ is hinged at ‘A’ from a wall and is held still with the help of a string, as shown in. This exercise involves drawing the free body diagram. To make the problem easier, the force F will be expressed in terms of its horizontal and vertical components. Removing all other elements from the image helps produce the finished free body diagram.

Free Body Diagram : The free body diagram as a finished product

Given the finished free body diagram, people can use their knowledge of trigonometry and the laws of sine and cosine to mathematically and numerical represent the horizontal and vertical components:

General Problem-Solving Tricks

Free body diagrams use geometry and vectors to visually represent the problem.

  • Construct a free-body diagram for a physical scenario

In physics, most problems are solved much more easily when a free body diagram is used. This uses geometry and vectors to visually represent to problem, and trigonometry is also used in determining horizontal and vertical components of forces and objects.

Purpose: Free body diagrams are very helpful in visually identifying which components are unknown, where the moments are applied, and help analyze a problem, whether static or dynamic.

How to Make A Free Body Diagram

To draw a free body diagram, do not worry about drawing it to scale, this will just be what you use to help yourself identify the problems. First you want to model the body, in one of three ways:

  • As a particle. This model may be used when any turning effects are zero or have zero interest even though the body itself may be extended. The body may be represented by a small symbolic blob and the diagram reduces to a set of concurrent arrows. A force on a particle is a bound vector.
  • rigid extended . Stresses and strains are of no interest but turning effects are. A force arrow should lie along the line of force, but where along the line is irrelevant. A force on an extended rigid body is a sliding vector.
  • non-rigid extended . The point of application of a force becomes crucial and has to be indicated on the diagram. A force on a non-rigid body is a bound vector. Some engineers use the tail of the arrow to indicate the point of application. Others use the tip.

Do’s and Don’ts

What to include: Since a free body diagram represents the body itself and the external forces on it. So you will want to include the following things in the diagram:

  • The body: This is usually sketched in a schematic way depending on the body – particle/extended, rigid/non-rigid – and on what questions are to be answered. Thus if rotation of the body and torque is in consideration, an indication of size and shape of the body is needed.
  • The external forces: These are indicated by labelled arrows. In a fully solved problem, a force arrow is capable of indicating the direction, the magnitude the point of application. These forces can be friction, gravity, normal force, drag, tension, etc…

Do not include:

  • Do not show bodies other than the body of interest.
  • Do not show forces exerted by the body.
  • Internal forces acting on various parts of the body by other parts of the body.
  • Any velocity or acceleration is left out.

How To Solve Any Physics Problem : Learn five simple steps in five minutes! In this episode we cover the most effective problem-solving method I’ve encountered and call upon some fuzzy friends to help us remember the steps.

free-body-diagram.png

Free Body Diagram : Use this figure to work through the example problem.

  • Dimensional analysis is the practice of checking relations amount physical quantities by identifying their dimensions.
  • It is common to be faced with a problem that uses different dimensions to express the same basic quantity. The following equation can be used to find the conversion factor between the two derived dimensions: \(\mathrm{n_2=\frac{u_2}{u_1} \times n_1}\).
  • Dimensional analysis can also be used as a simple check to computations, theories and hypotheses.
  • It is important to identify the problem and the unknowns and draw them in a free body diagram.
  • The laws of cosine and sine can be used to determine the vertical and horizontal components of the different elements of the diagram.
  • Free body diagrams use geometry and vectors to visually represent physics problems.
  • A free body diagram lets you visually isolate the problem you are trying to solve, and simplify it into simple geometry and trigonometry.
  • When drawing these diagrams, it is helpful to only draw the body it self, and the forces acting on it.
  • Drawing other objects and internal forces can condense the diagram and cause it to be less helpful.
  • dimension : A measure of spatial extent in a particular direction, such as height, width or breadth, or depth.
  • trigonometry : The branch of mathematics that deals with the relationships between the sides and the angles of triangles and the calculations based on them, particularly the trigonometric functions.
  • static : Fixed in place; having no motion.
  • dynamic : Changing; active; in motion.

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  • MyU : For Students, Faculty, and Staff

Jacquelyn Burt Earns 2024 John Tate Award for Excellence in Undergraduate Advising

Department of Computer Science & Engineering Undergraduate Academic Advisor Jacquelyn Burt was awarded the 2024 John Tate Award for Excellence in Undergraduate Advising. Named in honor of John Tate, Professor of Physics and first Dean of University College (1930-41), the Tate Awards serve to recognize and reward high-quality academic advising, calling attention to the contribution academic advising makes to helping students formulate and achieve intellectual, career, and personal goals.

“I thought it was a trick when I got the email that I was being nominated,” said Jacquelyn. “Within the advising field, this award is a big deal; I described it to my parents as ‘the advising Grammys’. Part of what makes it so cool is the nomination process, which involves several letters of support from students and colleagues as well as putting together a kind of portfolio of some of the programs and resources I’ve helped develop. So many different people contributed to that on my behalf, so it was really powerful to be reminded of the impact of my work and the amazing colleagues and students I get to love!”

Jacquelyn is a lifelong Gopher, earning her B.S. in business marketing education in 2014 and her M.Ed. in education policy and leadership in 2019. She joined the CS&E student services team in 2019, where she quickly developed a reputation as a staunch ally and advocate for her students. In 2021, Jacquelyn received the Gopher Spirit Award , recognizing the U of M advisor who contributes to a positive office culture, is inclusive, and brings others up. “I feel the most useful when a student or colleague is misunderstanding something, or experiencing a lot of stress, and I am able to help separate it into smaller pieces or come up with a different way of looking at it,” said Jacquelyn. “If I can shine light on something, help shift a lens or perspective, or give an idea or experience a bit of breathing room, I’m doing my job.”

When asked about what inspired her to work in advising, Jacquelyn replied, “When I first came to the University of Minnesota as a freshman, I was a family and social sciences major - I love relationships and helping, and so figured a career in marriage and family therapy sounded good. However, I’ve also always loved education and felt most at home at school - when I finished my undergraduate degree, I didn’t want to leave college because I loved it so much! Student advising seemed like a cool sweet spot between classroom teaching, advocacy, and being in a helping role. Ultimately, I’ve really come to see advising as facilitation work: I help students identify and navigate barriers to their goals, experiences, and personal development.”

As an undergraduate advisor, Jacquelyn manages a caseload of over 450 students in multiple majors, minors and other departmental programs. On top of her advising duties, Jacquelyn has undertaken a number of projects to better the undergraduate student experience, including establishing a weekly newsletter; designing, promoting, and executing departmental events and programs; and developing and teaching students through a variety of training and credit-bearing coursework. Most notably, Jacquelyn created and now facilitates mandatory implicit bias training for all 200+ undergraduate teaching assistants, as well as teaching CSCI 2915: Teaching Methods in Computer Science (a leadership and communication skills seminar) each semester.

“Within our student services team, we’ve developed a great culture of initiative and problem-solving: like, if you identify a problem and have or can create tools to help address it, amazing - you go get it!” said Jacquelyn. “We all believe that students deserve to have positive and supportive experiences while they are here, and we’ve built an advising team that trusts each of us to help bear that belief out. I definitely could not do my job without the collaboration, encouragement, and love of the whole team.”

On top of her work within CS&E, Jacquelyn has personally designed advising resources that have made an impact for undergraduate students across the entire university. Her “Explore & Expand” tool (originally developed for the college’s major/minor expo) is used widely throughout the entire University, particularly within the Center for Academic Planning and Exploration office. Additionally, her “Academic Progress Audit System Guide” resource (originally used within the departmental “Welcome to the Major” workshops) has been used in advisor training and onboarding. Above all, Jacquelyn has a keen eye for making connections, and for communicating things that can be overwhelmingly complex with both clarity and compassion.

“When I applied for this job, I had to come up with an ‘advising philosophy,’” said Jacquelyn. “What I landed on is anytime a student leaves an interaction with me, I want them to feel a little bit more seen, supported, and celebrated. I am a naturally celebratory person, which I’ve learned to embrace - and this award is a wonderful way to celebrate the work of advising!”

Learn more about the John Tate Award at the Provost website . 

Jacquelyn Burt headshot

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IMAGES

  1. Learn How to Solve Force Problems

    problem solving in force physics

  2. How to Solve Physics Problems with Forces

    problem solving in force physics

  3. PPT

    problem solving in force physics

  4. Guide to Solving Force Problems

    problem solving in force physics

  5. 6 Steps to Solve Force Problems

    problem solving in force physics

  6. Tension Force Physics Problems

    problem solving in force physics

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  1. 24.Physics

  2. Problem Solving-Force on continuous charge

  3. Chapter 9

  4. Ques- 9 to 12 FORCE Ex- 1(A) NUMERICALS PART

  5. STATICS LECTURE 2 Solving Force Vectors using the parallelogram law

  6. 5.8 equilibrium of a particle 5.9 common forces in mechanics ch-5 laws of motion

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  1. 6.1 Solving Problems with Newton's Laws

    Determine the system of interest. The result is a free-body diagram that is essential to solving the problem. Apply Newton's second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line. Check the solution to see whether it is reasonable.

  2. Solving problems which involve forces, friction, and Newton's ...

    Solving problems which involve forces, friction, and Newton's Laws: A step-by-step guide | Phyley Support Ukraine 🇺🇦 Help Ukrainian Army Humanitarian Assistance to Ukrainians In this tutorial you will learn how to examine and solve problems which involve forces and applications of Newton's Laws.

  3. Calculating Force

    Correct answer: The formula for force is. We are given the mass, but we will need to calculate the acceleration to use in the formula. We know the initial velocity (zero because the box starts from rest), final velocity, and distance traveled. Using these values, we can find the acceleration using the formula.

  4. Forces and Newton's laws of motion

    Unit 1 One-dimensional motion. Unit 2 Two-dimensional motion. Unit 3 Forces and Newton's laws of motion. Unit 4 Centripetal force and gravitation. Unit 5 Work and energy. Unit 6 Impacts and linear momentum. Unit 7 Torque and angular momentum. Unit 8 Oscillations and mechanical waves. Unit 9 Fluids.

  5. Net Force Problems Revisited

    Resolution of Forces Equilibrium and Statics Net Force Problems Revisited Inclined Planes Two-Body Problems This part of Lesson 3 focuses on net force-acceleration problems in which an applied force is directed at an angle to the horizontal.

  6. Newton's second law: Solving for force, mass, and acceleration

    Newton's second law: Solving for force, mass, and acceleration. A stunt woman of mass m falls into a net during the filming of an action movie. Assume she experiences upward acceleration magnitude a while touching the net.

  7. 4.5: Normal, Tension, and Other Examples of Forces

    Tension. A tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. The word "tension " comes from a Latin word meaning "to stretch.". Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons.

  8. 4.6: Problem-Solving Strategies

    Problem-Solving Strategy for Newton's Laws of Motion. Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure(a).

  9. 6.2: Forces

    Problem-Solving Strategy: Drawing Free-Body Diagrams. Draw the object under consideration. If you are treating the object as a particle, represent the object as a point. Place this point at the origin of an xy-coordinate system. Include all forces that act on the object, representing these forces as vectors.

  10. 1.7 Solving Problems in Physics

    Strategy Strategy is the beginning stage of solving a problem. The idea is to figure out exactly what the problem is and then develop a strategy for solving it. Some general advice for this stage is as follows: Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketch at the outset.

  11. 4.6 Problem-Solving Strategies

    Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 4.20 (a).

  12. What is tension? (article)

    Namely, we use Newton's second law to relate the motion of the object to the forces involved. To be specific we can, Draw the forces exerted on the object in question. Write down Newton's second law ( a = Σ F m) ‍. for a direction in which the tension is directed. Solve for the tension using the Newton's second law equation a = Σ F m.

  13. Vectors and Forces Problem Sets

    Problem 1: For each collection of listed forces, determine the vector sum or the net force. Audio Guided Solution Show Answer Problem 2: Hector is walking his dog (Fido) around the neighborhood. Upon arriving at Fidella's house (a friend of Fido's), Fido turns part mule and refuses to continue on the walk.

  14. Problem-Solving Strategies

    Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 1 (a).

  15. Forces in Physics, tutorials and Problems with Solutions

    SAT Questions on Forces with Solutions Physics Practice Questions with Solutions on Forces and Newton's Laws. Formulas and Constants Physics Formulas Reference SI Prefixes Used with Units in Physics, Chemistry and Engineering Constants in Physics, Chemistry and Engineering {ezoic-ad-1}

  16. 2.4: Problem Solving

    Example 2.4.3. A block is attached to one end of a massless spring, the other end of which is attached to a vertical fixed peg in a frictionless horizontal surface. The block is spun around a circle, and the spring stretches as a result of this motion. In fact, the faster the motion, the more the spring stretches.

  17. Newton's Law Problem Sets

    Problem 22: Brandon is the catcher for the Varsity baseball team. He exerts a forward force on the .145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m. Determine the acceleration of the ball and the force which is applied to it by Brandon.

  18. Force Problems

    Force Problems On this page I put together a collection of force problems to help you understand forces better. The required equations and background reading to solve these problems are given on the friction page, the equilibrium page, and Newton's second law page . Problem # 1 A ball of mass m is hanging on a wall with a string, as shown.

  19. Centripetal force problem solving (video)

    You divide by the radius which was 0.5, and you get that the force of tension had to be 100 Newtons. So in this case, the force of tension, which is the centripetal force, is equal to 100 Newtons. Now, some of you might be thinking, hey, this was way too much work for what ended up being a really simple problem.

  20. 8.1 Linear Momentum, Force, and Impulse

    We can solve for Δ p by rearranging the equation. F net = Δ p Δ t. to be. Δ p = F net Δ t . F net Δ t is known as impulse and this equation is known as the impulse-momentum theorem. From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts.

  21. Tension, String, Forces Problems with Solutions

    Problem 1 A block of mass 5 Kg is suspended by a string to a ceiling and is at rest. Find the force Fc exerted by the ceiling on the string. Assume the mass of the string to be negligible. Solution a) The free body diagram below shows the weight W and the tension T 1 acting on the block.

  22. Physics_ontable on Instagram: "Question:- Consider a wire bent in the

    237 likes, 13 comments - physics_ontable on February 3, 2024: "Question:- Consider a wire bent in the shape of a semi-circle with a radius of 2 meters. If a cur..." Physics_ontable on Instagram: "Question:- Consider a wire bent in the shape of a semi-circle with a radius of 2 meters.

  23. Kinematic Equations: Sample Problems and Solutions

    A useful problem-solving strategy was presented for use with these equations and two examples were given that illustrated the use of the strategy. Then, the application of the kinematic equations and the problem-solving strategy to free-fall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be presented.

  24. 1.4: Solving Physics Problems

    Trigonometry and Solving Physics Problems. In physics, most problems are solved much more easily when a free body diagram is used. Free body diagrams use geometry and vectors to visually represent the problem. Trigonometry is also used in determining the horizontal and vertical components of forces and objects.

  25. Jacquelyn Burt Earns 2024 John Tate Award for Excellence in

    Department of Computer Science & Engineering Undergraduate Academic Advisor Jacquelyn Burt was awarded the 2024 John Tate Award for Excellence in Undergraduate Advising. Named in honor of John Tate, Professor of Physics and first Dean of University College (1930-41), the Tate Awards serve to recognize and reward high-quality academic advising, calling attention to the contribution academic ...