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Check Your Understanding Suppose the radius of the loop-the-loop in [link] is 15 cm and the toy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop?

3 m/s

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Visit Carleton College’s site to see a video of a looping rollercoaster.

In situations where the motion of an object is known, but the values of one or more of the forces acting on it are not known, you may be able to use the work-energy theorem to get some information about the forces. Work depends on the force and the distance over which it acts, so the information is provided via their product.

Determining a stopping force

A bullet from a 0.22LR-caliber cartridge has a mass of 40 grains (2.60 g) and a muzzle velocity of 1100 ft./s (335 m/s). It can penetrate eight 1-inch pine boards, each with thickness 0.75 inches. What is the average stopping force exerted by the wood, as shown in [link] ?

In figure a, a bullet is moving horizontally at a speed of 335 meters per second toward a set of 8 boards, arranged in a horizontal stack. In figure b, the bullet has passed through the stack of boards and has stopped at the far end of the last board. The stopping distance is indicated as the width of the stack of boards.
The boards exert a force to stop the bullet. As a result, the boards do work and the bullet loses kinetic energy.

Strategy

We can assume that under the general conditions stated, the bullet loses all its kinetic energy penetrating the boards, so the work-energy theorem says its initial kinetic energy is equal to the average stopping force times the distance penetrated. The change in the bullet’s kinetic energy and the net work done stopping it are both negative, so when you write out the work-energy theorem, with the net work equal to the average force times the stopping distance, that’s what you get. The total thickness of eight 1-inch pine boards that the bullet penetrates is 8 × 3 4 in . = 6 in . = 15.2 cm .

Solution

Applying the work-energy theorem, we get

W net = F ave Δ s stop = K initial ,

so

F ave = 1 2 m v 2 Δ s stop = 1 2 ( 2.6 × 10 −3 kg ) ( 335 m/s ) 2 0.152 m = 960 N .

Significance

We could have used Newton’s second law and kinematics in this example, but the work-energy theorem also supplies an answer to less simple situations. The penetration of a bullet, fired vertically upward into a block of wood, is discussed in one section of Asif Shakur’s recent article [“Bullet-Block Science Video Puzzle.” The Physics Teacher (January 2015) 53(1): 15-16]. If the bullet is fired dead center into the block, it loses all its kinetic energy and penetrates slightly farther than if fired off-center. The reason is that if the bullet hits off-center, it has a little kinetic energy after it stops penetrating, because the block rotates. The work-energy theorem implies that a smaller change in kinetic energy results in a smaller penetration. You will understand more of the physics in this interesting article after you finish reading Angular Momentum .

Learn more about work and energy in this PhET simulation called “the ramp.” Try changing the force pushing the box and the frictional force along the incline. The work and energy plots can be examined to note the total work done and change in kinetic energy of the box.

Summary

  • Because the net force on a particle is equal to its mass times the derivative of its velocity, the integral for the net work done on the particle is equal to the change in the particle’s kinetic energy. This is the work-energy theorem.
  • You can use the work-energy theorem to find certain properties of a system, without having to solve the differential equation for Newton’s second law.
Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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