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Answer 1.1: a) The flour is leaking out at a rate of .04 lb/s

b) Hal will do 123 ft-lbs of work

c) The bag will empty before he reaches the top, so he will only end up doing 69.4 ft-lbs of work

Answer 1.2: You would do 16.6 J more work

Answer 1.3: a) Luisa does 532 kJ of work, Miguel does 684 kJ

b) No, they actually did more work because the forces they exerted were nonconservative. The amount of work they actually did depends on the path taken, their speed at each point, and their metabolic rates.

c) From the chemical energy in their food

d) It will become heat, making them sweaty and out of breath

Answer 1.4: a) Sam does 2144 J of work

b) Cory is moving at 9.24 m/s at the bottom

Answer 1.5: Because the direction of the force changes with the distance the roller travels, we need to take the integral of the force dotted with the distance over the distance and end up with 40.2 J


Answer 2.1: a) 449 J

b) It is moving at 8.47 m/s

Answer 2.2: a) 1.41 x 1011 J

b) 2.87 x 1011 J; since her distance is almost 10% of the earth’s radius, the gravitational field strength is noticeably reduce

Answer 2.3: a) P=E/t and E=mgh, so water goes over the dam at a rate of1.93 x 106 kg/s

b) 46.7 m/s

Answer 2.4: a) .859 m/s

b) 388 m/s

Answer 2.5: a) 81.75 N/m

b) 501 J

Answer 2.6: a) 450 J

b) His maximum speed is when the net force on him is equal, when the cord has stopped accelerating him and friction is not yet slowing him down, and at that point his speed is 2.65 m/s

c) 4.1 m

d) 450 J


Answer 3.1: a) 816 J

b) 4.2 m/s

c) 7.8 m/s

Answer 3.2: a) .0566 J

b) .00579o Celsius

Answer 3.3: a) 4.85 m/s

b) .295 m

c) If the monkeys drop the bananas at the top of their swing, it won’t affect anything about their swing back down. However, the bananas will probably be stolen by someone else before they get back down.

Answer 3.4: a) 2 m vertically above the lowest point

b) 6.26 m/s

c) 462 J

Answer 4.1: 1.6 x 104 m3

Answer 4.2: a) 78.2 g

b) 210 o F

c) No, mass is so small

Answer 4.3: a) 24 photons at 690 nm is 1.38 x 10-17 J of energy

b) 38.6 %


Answer 5.1: a) 15 s

b) Using pulleys means that Joe won’t have to pull as hard because the ropes divide the load. Unfortunately, it also means that he has to walk four times as far to raise the water from the well. So the power (rate at which he does work) is smaller, but he still has to do the same amount of work.

Answer 5.2: a) 3.17 x 1013 kg of coal, 1.16 x 1014 kg of CO 2

b) 9000 kg of enriched uranium

Answer 5.3: a) 40 cells

b) About 1 yr, 10 mos

Answer 5.4:

Answer 5.5:

Answer 5.6:

where r is the density of the air. The kinetic power of the wind depends on the cube of the wind velocity, so a small increase in the wind speed will slow the ball down significantly.

Answer 5.7: The power in the wind is
and 40% of this is converted into electricity by the turbine. We solve this equation for A to get:

The area swept out by the turbine blades is a circle, so the radius of the turbine will be 3.9 m.

Answer 5.8: We assume that the wind is steady in direction and magnitude, that the density of the air is constant, and that the flow is uniform and incompressible. That means that we can assume the air flows through the turbine in a streamline, as shown above, which means that the flow of mass through the turbine is:

These assumptions also mean that Bernoulli’s equation holds:

Ï b + r v b 2 = Ï a + r v a 2

In the two equations above, r is the density of the air, A b, A t, and A a are the cross-sectional area of the streamline before, at, and after the turbine, respectively; v b, v t, and v a are the speed of the air before, at, and after the turbine, respectively; and Ï b and Ï a are the air pressure before and after the turbine, respectively.

First consider the force on the rotor, and conservation of momentum to derive:

Further, consider force as a difference of pressure, and use Bernoulli’s equation to derive:

F=A­ t ( Ï a- Ï b) = r A t (v b 2-v a 2)

Setting these two equal, show that the speed of the wind at the turbine is an average of the speeds before and after:

Now, the power extracted by the turbine is equal to the difference in the wind power before and after the turbine. Using the equation we derived in problem 1 for the kinetic power of the wind:

we can find:

For a given v b, we want to find the v a that maximizes P t. Show that this occurs for v a= v b/3. Substitute this back in to find:

The quantity in parentheses is the power of the wind impinging on the turbine, so the turbine can extract 16/27 = 59% of the power of the wind. This is known as the Betz Limit.

So Mariella is right, and Ignacio is unnecessarily pessimistic about the potential of wind power.

Answer 5.9: a) The turbine/generator converts into electricity 45% of the power in the wind, so

and A = 31.6 m 2. The area swept out is circular, so A = p r 2 and r = 3.2 m.

b) We reuse the same power equation, substituting different windspeeds, and get:

7 W at v = 1 m/s

57 W at v = 2 m/s

7 kW at v = 10 m/s

Note the very strong dependence on windspeed.

Answer 5.10:

Answer 5.11:

Answer 5.12:

Answer 5.13:


Energy Problems


Posted on 8/1/05

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