Work and Energy Name ______________________________

Work = Force ´ Distance P.E. = Weight ´ Height K.E = ½ Mass ´ Velocity ²

When solving these problems, use the following units:

Mass = kilograms

Work = Joules

Force = Newtons

Distance = meters

P.E. = Joules

Weight = Newtons (weight is the FORCE of gravity on an object)

Height = meters

K.E. = Joules

Velocity = meters/second

A man lifts bowling ball weighing 54 _(unit) over his head, and holds it there for 5 minutes. In lifting the ball he gave it 124 _(unit) of energy.

1. What unit would be used to measure the weight of the bowling ball? Is it Joules, Kilograms, or Newtons?

2. What unit would be used to measure the energy gained by the bowling ball? Is it Joules, Kilograms, or Newtons?

3. How high did he lift the bowling ball? Use the equation: P.E. = Weight ´ Height and solve it for height

4. How much work did he do on the bowling ball during the 5 minutes he held it up? Zero.

Since the distance the force was applied = zero, so does the amount of work done

A baseball bat applies 36 N of force to a baseball. It is in contact with the ball across a distance of 0.76 meters.

5. How much energy does the bat transfer to the ball?

Work = Force ´ Distance

= 36 N ´ 0.76 m

6. If the ball was hit horizontally, what form would the energy gained by ball take?

(see listing of forms of energy) Hint: all of its energy went into changing its motion

7. If the ball was hit vertically, and the ball weighed 1.4 N, how high could the ball go?

P.E. = Weight ´ Height

8. If the ball was hit horizontally, and left the bat at a velocity of 19.6 m/s, what is the

mass of the ball?

Solve the equation K.E = ½ Mass ´ Velocity ² for mass.

A stretched rubber band shoots a paper “bullet” having a mass of .0004 kg at a velocity of 43 m/s.

9. How much energy was stored in the stretched rubber band?

K.E = ½ Mass ´ Velocity ²

10. What form of energy was present in the stretched rubber band?

(hint: it was a form of stored enrgy)

11. If, in stretching the rubber band, it was pulled back a distance of 0.22 meters, what was the average

force needed to stretch it?

Use the equation Work = Force ´ Distance

Use the value from #9 for work (remember, work is energy) and 0.22 m for the distance

A force of 668 N is needed to push a box across a carpeted floor, gaining 4940 J of energy.

12. How far was the box pushed?

4940 J = 668 N ´ Distance

4940 J / 668N = Distance

7.39 meters

13. If the box did not end up gaining any velocity, what form did the energy take? Think friction (try rubbing your hands together)

A battery containing 2455 J of energy is used to power a device that lifts a 12 N rock up the side of a building.

14. How high could it lift the rock?

P.E. = Weight ´ Height (remember energy is energy, no matter what form it started out as)

15. what form of energy was present in the battery? Chemical potential

Answers:

0 J

0.14 kg

0.37 J

1.7 N

2.3 m

7.4 m

19 m

27 J

204 m