Work, Power, & Energy

### SAMPLE PROBLEMS

1. Dino Dude pulls his 7 kg wagon 5 m up a hill (at 40 from level) at a constant acceleration of 0.4 m/s2. How much work is done on the wagon? How much work does Dino Dude do? If it takes him 4 seconds to cover the distance, how much power does Dino Dude exert? If Dino Dude burns 250 calories as he pulls the wagon up the hill, how efficient are his movements? (one calorie is equal to 4.18 J)

First draw a free body diagram (B.) and list givens and unknowns:

m = 7 kg
d = 5 m
t = 4 s
Ein = 225 cal
Wnet (hill) = ?
Wboy (hill) = ?
Pboy = ?
Efficiency = ?

Note: Since the wagon accelerates up the hill, that is the direction of the net force acting on the wagon and the direction of the work done. Therefore, in answering this question we must focus on the components of work and force vectors which are parallel to the surface of the hill.

Proceed to find one unknown at a time...

Wnet hill = ?

In order to find Wnet hill we must first find Fnet hill.

Fnet hill = ma = 7 kg x 0.4 m/s = 2.8 N
Wnet hill = Fnet hill x d = 2.8 N x 5 m = 14 J
Wnet hill = 14 J

Wboy hill = ?

In order to find Wboy hill we must first find Fboy hill. From the free-body diagram we see that Fnet hill = Fboy hill + Fgravity hill.

Fboy hill = Fnet hill - Fgravity hill
Fgravity hill = Fgravitycos = 7 kg x (-9.81 N/kg) x cos50 = -44.14 N
Fboy hill = Fnet hill - Fgravity hill = 2.8 N - (-44.14 N) = -46.94 N
Wboy hill = Fboy hill x d = -46.94 N x 5 m = 234.7 J
Wboy hill = 234.7 J

Pboy = ?

Pboy hill = Wboy hill / t = 234.7 J / 4 s = 58.68 watts
Pboy hill = 58.68 watts

Efficiency = ?

Ein = 250 calories x 4.18 J/ calorie = 1045 Joules
Eout = Wboy hill = 234.7 J
Efficiency = Eout / Ein = 234.7 J / 1045 J = 0.22

Efficiency = 22%

2. Dino Dude is peacefully floating over the water when a big blue bird pops his balloons, causing Dino Dude to fall 10 m before crashing into the water. If Dino Dude has a mass of 20 kg, what is his velocity immediately before impact with the water? How much mechanical work is required to stop Dino Dude just before he impacts with the water?

The principle of conservation of mechanical energy must be used to solve the first part of this problem. The total mechanical energy Dino Dude possesses at a height of 10 m is his potential energy. Immediately before impact with the water, his gravitational potential energy can be assumed to be 0 and 100% of his mechanical energy will then be kinetic.

a) Find the total energy possessed by Dino Dude

PE + KE = Constant
magh + 1/2 mv2 = Constant
(20 kg)(9.81 m/s2)(10 m) + 1/2 (20 kg)(0)2 = Constant
1962 J = Constant

b) Find Dino Dude's velocity just before impact

PE + KE = 1962 J
magh + 1/2 mv2 = 1962 J
(20 kg)(9.81 m/s2)(0 m) + 1/2 (20 kg)(v2) = 1962 J
v2 = 196.2 m2/s2 v = 14.0 m/s

The principle of work and energy must be used to calculate the work required to change Dino Dude's kinetic energy to 0. Assume that the potential energy and thermal energy of Dino Dude do not change.

c) Find mechanical work required to stop Dino Dude

W = KE
W = (1/2 mv2)f - (1/2 mv2)i W = 0 - 1/2 (20 kg)(14.0 m/s)2
W = 1960 J