**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/s_{2}. 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

E_{in} = 225 cal

W_{net (hill)} = ?

W_{boy (hill)} = ?

P_{boy} = ?

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...

W_{net hill} = ?

In order to find W_{net hill} we must first find F_{net hill}.

F_{net hill} = ma = 7 kg x 0.4 m/s = 2.8 N

W_{net hill} = F_{net hill} x d = 2.8 N x 5 m = 14 J

**W**_{net hill} = 14 J

W_{boy hill} = ?

In order to find W_{boy hill} we must first find F_{boy hill}.
From the free-body diagram we see that F_{net hill} = F_{boy
hill} + F_{gravity hill}.

F_{boy hill} = F_{net hill} - F_{gravity hill}

F_{gravity hill} = F_{gravity}cos = 7 kg x (-9.81
N/kg) x cos50 = -44.14 N

F_{boy hill} = F_{net hill} - F_{gravity hill}
= 2.8 N - (-44.14 N) = -46.94 N

W_{boy hill} = F_{boy hill} x d = -46.94 N x 5 m = 234.7
J

**W**_{boy hill} = 234.7 J

**P**_{boy} = ?

P_{boy hill} = W_{boy hill} / t = 234.7 J / 4 s = 58.68
watts

**P**_{boy hill} = 58.68 watts

Efficiency = ?

E_{in} = 250 calories x 4.18 J/ calorie = 1045 Joules

E_{out} = W_{boy hill} = 234.7 J

Efficiency = E_{out} / E_{in} = 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

ma_{g}h + 1/2 mv^{2} = Constant

(20 kg)(9.81 m/s^{2})(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

ma_{g}h + 1/2 mv^{2} = 1962 J

(20 kg)(9.81 m/s^{2})(0 m) + 1/2 (20 kg)(v^{2}) = 1962
J

v^{2} = 196.2 m^{2}/s^{2} **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 mv^{2})_{f} - (1/2 mv^{2})_{i}
W = 0 - 1/2 (20 kg)(14.0 m/s)^{2}

**W = 1960 J **

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