1. calculate the work done by a 37 n force pushing a pencil 0.46 m.
2. calculate the work done by a 57 n force pushing a 0.025 kg pencil 0.25 m against a force of 25 n.
3. calculate the work done by a 2.7 n force pushing a 425 g sandwich across a table 0.95 m wide.
4. how far can a mother push a 20.0 kg baby carriage, using a force of 66 n, if she can only do 2930 j of work?
5. how much work is it to lift a 15 kg sack of potatoes vertically 5.5 m?
6. if a small motor does 620 j of work to move a toy car 270 m, what force does it exert?
7. a girl pushes her little brother on his sled with a force of 290 n for 780 m. how much work is this if the force of friction acting on the sled is (a) 210 n, (b) 310 n?
8. a 78.0 kg man pushes on a 500,000 t wall for 250 s but it does not move. how much work does he do on the wall?
9. a boy on a bicycle drags a wagon full of newspapers at 0.95 m/s for 35 min using a force of 48 n. how much work has the boy done?
10. what is the gravitational potential energy of a 65.2 kg person standing on the roof of a 15-storey building relative to (a) the tenth floor, (b) the sixth floor, (c) the first floor. (each storey is 2.50 m high.)
11. a 1. x 10^4 kg airplane lands, descending a vertical distance of 15 km while travelling 150 km measured along the ground. what is the plane's loss of potential energy?
12. a coconut falls out of a tree 15.0 m above the ground and hits a bystander 3.00 m tall on the top of the head. it bounces back up 1.60 m before falling to the ground. if the mass of the coconut is 2.70 kg, calculate the potential energy of the coconut relative to the ground at each of the following sites:
(a) while it is still in the tree,
(b) when it hits the bystander on the head,
(c) when it bounces up to its maximum height,
(d) when it lands on the ground,
(e) when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole.

Question1 The work of the force on that body will be equivalent to the energy necessary to displace it in an accelerated manner. The work is a scalar physical magnitude that is represented by the letter (of the English Work) and is expressed in units of energy, this is in joules or joules (J) in the International System of Units.
 W = F * d
 W = (37N) * (0.46m) = 17.02Nm
 W = 17.02 J
 Answer 
 17.02J 

 Question 2
 Using the work definition, we have the following equation:
 W = F * d
 For this case we must make a free body diagram in the direction x and find the resultant force.
 F = (57N-25N) = 32N
 Then, the work done will be
 W = F * d
 W = (32) * (0.25) = 8Nm
 W = 8J
 Answer
 8J

 Question 3
 We use the work definition, where in this case we have the force in the direction of the displacement is 2.7N and the distance traveled is 0.95m. 
 Then, the work done will be:
 W = F * d
 W = (2.7) * (0.95) = 2.565Nm
 W = 2.565J
 Answer
 2.565J

 Question 4 
 For this case we must use the definition of work to find the distance that the mother can travel when pushing the baby's car.
 W = F * d
 Clearing we have
 d = W / F
 Substituting the values and remembering that J equals Nm, we have
 d = (2930) / (66) = 44.39m
 d = 44.39m
 Answer
 the mother can push the car a distance of 44.39m

 Question 5
 For this case we must find the work done in by force in the vertical direction. 
 For this, we make a free body diagram in which we have the following:
 F = mg = (15) * (9.8) = 147N
 Then, if we can use the work definition:
 W = F * d
 Substituting:
 W = (147) * (5.5) = 808.5Nm
 W = 808.5J
 Answer
 808.5J

 Question 6
 For this question, the first thing you should see is that in the definition of work, you have two variables that are data. The first variable is the work done by the engine which is 620J and the second variable is the distance traveled by the car which is 270m.
 Using the definition, we have
 W = F * d
 Clearing we have
 F = W / d
 Substituting and remembering that J equals NM
 F = (620) / (270) = 2.30 N
 F = 2.30 N
 Answer
 2.30 N

 Question 7
 For this case we must make a free body diagram in the direction x to find the resultant force and then be able to find the work done.
 For (a) 210 N
 F = 290-210 = 80N
 Then, the work will be
 W = F * d
 replacing:
 W = (80) * (780) = 62400J
 For (b) 310 N
 F = 290-310 = -20N
 Then, the work will be
 W = F * d
 replacing:
 W = (- 20) * (780) = - 15600J
 answer
 W = 62400J
 W = -15600J

 Question 8

 For this case we must first remember the definition of work which is
 W = F * d
 Since there is no distance to travel since the wall does NOT move then
 d = 0
 Therefore the work done is
 W = 0J
 answer
 W = 0J

 Question 9
 For this case you must remember two definitions:
 Distance equals speed by time.
 Work equals force by distance traveled.
 Using the definitions we have
 d = v * t
 d = (0.95) * (35 * 60) = 1995m
 Then the work is
 W = F * d
 W = (48) * (1995) = 95760Nm
 W = 95760J
 answer
 95760J.

 Question 10
 For this case, let's see the definition of gravitational potential energy.
 U = m * g * h
 Where
 m = mass
 g = severity
 h = height
 (a) the tenth floor 
 U = (65.2) * (9.8) * (2.50 * 10) = 15974N
 U = 15974N
 (b) the sixth floor
 U = (65.2) * (9.8) * (2.50 * 6) = 9584.4N
 U = 9584.4N
 (c) the first floor
 U = (65.2) * (9.8) * (2.50) = 1597.4N
 U = 1597.4N
 Answer
 15974N
 9584.4N
 1597.4N

   
 Question 11
 It can be said that the potential energy of the plane is
 U = (10000) * (9.8) * (15000) = 1.47 * 10 ^ 9J
 However, there is no loss of potential energy since this energy is then transformed into kinetic energy.
 The energy is not lost.
 Answer
 1.47 * 10 ^ 9J are transformed into kinetic energy. The energy is not lost.

 Question 12
  We must use the definition of potential energy for each of the cases:
 (a) while it is still in the tree, 
 U = mgh
 U = (2.70) * (9.8) * (15) = 396.9J
 U = 396.9J
 (b) when it hits the bystander on the head,
 U = mgh
 U = (2.70) * (9.8) * (15-3) = 317.52J
 U = 317.52J
 (c) when it bounces up to its maximum height, 
 U = (2.70) * (9.8) * (1.60) = 42.336J
 U = 42,336J

 (d) when it lands on the ground,
 U = (2.70) * (9.8) * (0) = 0J
 U = 0J
 (e) when it rolls into a groundhog hole, and falls 2.50 m to the bott

 U = (2.70) * (9.8) * (- 2.50) = - 66.15J
 U = -66.15J
 Answer
 U = 396.9J
 U = 317.52J
 U = 42,336J
 U = 0J
 U = -66.15J
 Note how the potential energy decreases with height.