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**Work** changes the energy of a system. If the only change resulting from applying a force on an object is to change the position of the object, then the work done on the object is the change in **gravitational potential energy**. The path of the object does not matter; the force on the object does not have to be constant.

The **gravitational potential energy** of an object near the surface of a planet is mgh, where m is the mass of the object, g is the gravitational field intensity, and h is the height of the object.

1. How much work is done by a crane lifting a 200.0 kg crate from the ground to a floor 21.0 m above the ground. What is the change in gravitational potential energy of the crate?

2. A 25-kg box slides, from rest, down a 9.0-m-long incline that makes an angle of 15° with the horizontal. The speed of the box when it reaches the bottom of the incline is 2.4 m/s.

a. What is the coefficient of kinetic friction between the box and the surface of the incline?

b. How much work is done on the box by the force of friction and

c. What is the change in the potential energy of the box?

3. A 40.0-kg wagon is towed up a hill inclined at 18.5º with respect to the horizontal. the tow rope is parallel to the incline and has a tension of 140N in it. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. How fast is the wagon going after moving 80 m up the hill?

4. A 25.6kg child pulls a 4.81kg toboggan up a hill inclined at 25.7° to the horizontal. The vertical height of the hill is 27.3 m. Friction is negligible. Determine how much work the child must do on the toboggan to pull it at a constant velocity up the hill.

5. An 81.0-kg in-line skater does +3500 J of nonconservative work by pushing against the ground with his skates. In addition, friction does -710 J of nonconservative work on the skater. The skater's initial and final speeds are 2.50 m/s and 1.60 m/s, respectively. (a) Has the skater gone uphill, downhill, or remained at the same level? (b) Calculate the change in height of the skater.

6. A solid object with mass (m) is initially at rest. An applied constant vertical force (F) causes the object to reach an upward speed (V), and total displacement (h). Use Newton's Second Law to derive an expression (in terms of m,g,h, & V) for the work.

Selected solutions are printed below.

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1. W = mgh_{final} - mgh_{initial}

= (200.0 kg)(9.8 N/kg)(21.0) - (200.0 kg)(9.8 N/kg)(0)

= 41160 J

Change in gravitational potential energy = 41160 J

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