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**Work** changes the **energy** of a system. When the energy of the system increases, work is positive. The work done on a rigid body by a net force changes the energy of the body. The path of the object does not matter; the force on the object does not have to be constant.

If the only change resulting from applying a force on an object is change in velocity, then the work done on the object is the change in **kinetic energy**.

W = E_{kfinal} - E_{kinitial} = (0.5)m(v_{final})^{2} - (0.5)m(v_{initial})^{2}

When a uniform force is applied to a rigid body moving in a straight line,

W = Fdcosø

(See bottom of page for answers.)

1. A boy pushes a 5.00 kg cart in a circle, starting at 0.500 m/s and accelerating to 3.00 m/s. How much work was done on the cart?

2. A 30.0 kg box initially sliding at 5.00 m/s on a rough surface is brought to rest by 20.0 N of friction. What distance does the box slide?

3. A 1000.0 kg truck accelerates from 20.0 m/s to 25.0 m/s over a distance of 300.0 m. What is the average net force on the truck?

4. A space ship of mass 5.00 ×10^{4} kg is traveling at a speed 1.15 × 10^{4} m/s in outer space. Except for the force generated by its own engine, no other force acts on the ship. As the engine exerts a constant force of 4.00 × 10^{5} N, the ship moves a distance of 2.50 × 10^{6} m in the direction of the force of the engine.

a. Determine the final speed of the ship using the work-energy theorem.

b. Determine the final speed of the ship using the kinematics equations.

5. A force of 6.0 N is used to accelerate a mass of 1.0 kg from rest for a distance of 12m. The force is applied along the direction of travel. The coefficient of kinetic friction is 0.30. What is the

a. work done by the applied force?

b. work done by friction?

c. kinetic energy at the 12-m mark?

6. A 0.600-kg particle has speed of 2.00 m/s at point A and kinetic energy of 7.50 J at point B. What is

a. its kinetic energy at A?

b. its speed at B?

c. the total work done on the particle as it moves from A to B?

7. Two carts, one twice the mass of the other, experience the same force for the same time. What is their difference in momentum? What is their difference in kinetic energy?

8. A weapon fired a 25.8-kg shell with a muzzle speed of 880 m/s. What average force acted on the shell?

9. A catcher stops a 91 mi/h pitch in his glove, bringing it to rest in 0.00179 m. If force exerted by the catcher is 785 N, what is the mass of the ball?

10. A 7.80 kg package was dropped onto a flatbed moving horizontally at 1.60 m/s. The coefficients of friction are as follows: µ_{s}= 0.470 and µ_{k }= 0.150 . How far does the package slide on the flatbed?

11. A 12.4 g bullet is fired horizontally into a 96 g wooden block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.650, what was the speed of the bullet immediately before impact?

12. A toy cart moves with a kinetic energy of 30 J. If its speed is doubled, what is its kinetic energy?

13. A car traveling 59 miles/hour locks its brakes until the car reaches 42 miles/hour. The mass of the car is 79.2 slugs. Calculate the energy lost to friction.

14. A 70 kg diver steps off a 10 m tower and drops from rest straight down into the water. If he comes to rest 5 m beneath the surface, determine the average force exerted by the water.

15. A 1300-kg car slows from 18 m/s to 15 m/s through a distance of 30 m. (a) Was the net work done on the car positive, negative, or zero? Explain. (b) Find the magnitude of the average net force on the car in the sandy section.

Selected solutions are printed below.

For solutions to all the problems on this page click here.

1. W = E_{kfinal} - E_{kinitial} = (0.5)m(v_{final})^{2} - (0.5)m(v_{initial})^{2}

W = (0.5)(5.00)(3.00)^{2} - (0.5)(5.00)(0.500)^{2}

W = = 21.9 J

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