# Energy and Work (others)

#### Weatherkid11

Please let me know if I am on the right track for these problems. There are two other ones i have added. PLease let me know if i am right for all four. Thanks a bunch!
1) A weight lifter picks up a barbell and
1. lifts it chest high
2. holds it for 5 minutes
3. puts it down.
Rank the amounts of work W the weight lifter performs during this three operations. Label the quantities as W1, W2, and W3. Justify you ranking order. ---> I think that the most amount of work is when it is lifted chest high, since there is a force and displacement involved. After that, would me when he puts it down, becuase gravity helps, and there is a displacement. When he holds it up for five min there is no work becuase there is no displacement.
2)Estimate the amount of work the engine performed on a 1200 kg car as it accelerated at 1.2 m/s2 over a 150 meter distance. ----> Well since the equation for work is force times distance, it would be (1.2 m/s2)(1200kg)(150m) So the total work would be 216,000 J. But this seems to easy, so i think i am wrong
3) Image for this promblem: http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/chapter7/medialib/chap7gifs/chap7_wu_4.gif [Broken]
A real world roller-coaster released at point A and coasting without external power would traverse a track somewhat like the figure above. Friction is not negligible in the real world. If we call the potential energy at point A relative to the ground Uo what can you say about the potential and kinetic energies UB, UB at point B and the potential and kinetic energies UC, UC at point C.----> Here is what I think: KA + UA = ( KB + UB ) + ( KC + UC ) + Ffr
So plugging in the variables, we get: mg(ha) = ( 1/2mvb^2 + mg(hb)) + ( 1/2mvb^2 + mg(hc))+Ffr
where:
m=mass
vb= velocity at point B
g= gravity
hb= height of point B
hc= height of point C
Ffr= Frictional force
4) A gallon of gasoline contains about 1.3 x 10^8 joules of energy. A 2000 kg car traveling at 20 m/s skids to a stop. Estimate how much gasoline it will take to bring the car back to the original speed? To complicate matters further, consider the fact that only about 15% of the energy extracted from gasoline actually propels the car. The rest gets exhausted as heat and unburnt fuel. ---> Wow... this one i am totally lost. Please help

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#### Diane_

Homework Helper
*Whew* Lotsa questions.

1) One of the problems with the way "work problems" are often stated is that they ignore the fact that work is not just "done" - it's done on something or by something. Assuming we're talking about the work done on the barbell, you've got it.

2) I started to say no, but the key word here is "estimate". The force in the work equation is the force that's causing the displacement. If you were sitting in a car pushing on the dashboard while the car hurtled down the Autobahn, you would be doing no work on the car even though there is a force and a displacement. The two are unrelated - multiplying them together makes no sense.

What you've done is taken the mass of the cart and multiplied by its acceleration - essentially, you've found the net force on it. Multiplying by the distance gives you the work done by the net force. This is clearly an underestimate for the work done by the engine, as it's also having to overcome friction in various different places, and friction in a car accelerating like this will be considerable. However, given the information you have, I don't see that you could make a better estimate.

3) There are a couple of problems here. First off, the equation you wrote would have you adding energies (your KB, UB, KC and UC) to forces. The dimensions are incompatible.

Secondly, your initial conditions KA + UA will give you the total mechanical energy for the problem. So far, so good. However, you equate it to the sum of all of the energies at all of the other points in the problem. It's like saying I have $300 in the bank right now. At noon, I have$250. At 1PM, I have $200. At 2PM, I have$150. So 300 = 250 + 200 + 150. This is clearly wrong, because the right side of my equation is double or triple adding the things that have been in my account all along. Does this make sense?

Personally, I wouldn't try for an equation on this problem. A qualitative description of the process would, I think, be a better approach. If you're interested, you might google for something called a "friction slope". People who design roller coasters use that as a basic tool.

4) There's a certain amount of energy in some particular form in the car when it's moving. When it's stopped, that energy is gone and will need to be replaced to get it back up to speed again. Where will that energy come from? Ignore the 15% at first and get an answer as though it were 100% efficient. When you have that answer, it'll be easier to put the 15% back in. If you don't see how, try it as though it were 50% efficient - what would that mean about the amount of gas you'll need as compared to the 100% case?

Hope this helps.

#### daniel_i_l

Gold Member
3) The total energy at point A is the total energy for the problem. The total mechanical energy at point B is equal to that at point A, but if you also look at the friction the the mechanical energy at B is a little less than that of point A because some of A's energy is wasted on friction (It is turned into heat)
The same is true for C relitive to B.

It is important to note that your Ffr is not constant during the trip, because the slope isn't constant so the N also isn't

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