MamaMia!
I thought I understood the curriculum, but when it came time to apply the formulas I drew a blank.

This is the question from the text. Any information you could offer would be greatly appreciated. Thanks!

A sled and rider with a combined mass of 50 kg are at the top of a hill a height of 15 m above the level ground below. The sled is given a push providing an initial kinetic energy at the top of the hill of 1600 J.

a. Choosing a reference level at the bottom of the hill, what is the potential energy of the sled and the rider at the top of the hill?

b. After the push, what is the total mechanical energy of the sled and the rider at the top of the hill?

c. If friction can be ignored, what will be the kinetic energy of the sled and the rider at the bottom of the hill?

I think the formula for a is PE=mgh. 7350 J.
I think the formula for b is PE + KE = 8950.
I think that c should = zero, but I'm not sure how. (Or is it W = -fd?)

Many thanks!
rpenner
QUOTE (MamaMia!+Jul 8 2006, 10:30 PM)
c.  If friction can be ignored, what will be the kinetic energy of the sled and the rider at the bottom of the hill?

I think that c should = zero, but I'm not sure how. (Or is it W = -fd?)

You are attempting to apply the wrong principle.

Since no forces are stated, and the shape of the hill is unknown, the work equation (which is change of energy = force times distance time cosine of angle between them) cannot be applied.

The lack of friction, means no work is being done on the sled by the ground, so the sled and rider are only acted on by gravity.

So for ©, KE(start) + PE(start) = KE(end) + PE(end)
but PE(end) = m * g * 0 = 0, so KE(end) = KE(start) + PE(start) = 8950 J, assuming g = 9.8 m/s^2 as you did in part (a)
MamaMia!
QUOTE (rpenner+Jul 8 2006, 10:59 PM)
You are attempting to apply the wrong principle.

Since no forces are stated, and the shape of the hill is unknown, the work equation (which is change of energy = force times distance time cosine of angle between them) cannot be applied.

The lack of friction, means no work is being done on the sled by the ground, so the sled and rider are only acted on by gravity.

So for ©, KE(start) + PE(start) = KE(end) + PE(end)
but PE(end) = m * g * 0 = 0, so KE(end) = KE(start) + PE(start) = 8950 J, assuming g = 9.8 m/s^2 as you did in part (a)

[QUOTE]

I was thinking in terms of acceleration....I am having a hard time getting my mind around the different concepts. I do understand it. That was actually the answer I had, but I thought it should be zero because the sled had stopped. As I said, I was thinking in terms of acceleration.

Thank you so much.

Also, if you don't mind could you answer the following:

Suppose that work is done on a large crate to tilt the crate so that it is balanced on one edge rather than sitting squarely on the floor as it had been previously. Has the potential energy of the crate increased in this process?

I think it has because the crate has changed positions and was moved against gravity, an opposing force.
rpenner
QUOTE (MamaMia!+Jul 8 2006, 11:31 PM)

Assuming g=9.8 meters/second^2.
You will get more points for correctness for "showing your work" which consists of showing the equations you are using, the substitutions of variables and the results of calcuations.
The shortest statement you could write for part (a) and expect to get full credit is
PE(top of hill) = mgh = (50 kg)(9.8 m/s^2)(15 m) = 7350 J
That way, you get partial credit if you forget to carry the one.
Likewise, for part ( b ) ,
E = KE + PE = 1600 J + 7350 J = 8950 J
QUOTE (MamaMia!+Jul 8 2006, 11:31 PM)
Suppose that work is done on a large crate to tilt the crate so that it is balanced on one edge rather than sitting squarely on the floor as it had been previously. Has the potential energy of the crate increased in this process?

I think it has because the crate has changed positions and was moved against gravity, an opposing force.

Assuming the box is a rectangular block, LxWxH, of constant densit, rho, then we can figure it's change in potential energy from the height of it's center of mass. The mass of the box is m, which is rho times L times W times H.

CODE
|    ___________
|   /\          |
|  /  H         H cos theta
| /    \        |
|+  CM  +--------
| \    /        |
|  \  L         L sin theta
|___\/__________|

Normally, the height of it's center of mass is h = 1/2 H. Assuming L > H, then as you tilt the box up to an angle, theta, the new height of the center of mass is h = (H cos(theta) + L sin(theta))/2, so the change in potential energy is

change in PE = mg(H cos(theta) + L sin(theta) - H)/2

Naturally, if L < H and theta is close to a right angle, then the change in PE is going to be negative, because the PE is smallest when the box is in the position where it has the smallest height.
MamaMia!
That was a great explanation! I had forgotten include that it was moved farther from the center of the earth. Your last sentence drove the point home. Your explanation and instruction is getting the synapses firing. Thank you!

I know this seems trivial for you, but it is setting the foundation of my understanding of physics. I truly appreciate you taking the time to reply.

The 'good turn' will come back to you - believe that!

Thanks again. Granny is learning - slow but sure!
Moseley
Welcome to the forum, in future you might place any such queries in the Homework Help section, being as it has little to do with Relativity, QM etc, although rpenner's expertise should still be available.
You may also come across previous posts which can assist you, a large number of the equations you are likely to need are discussed in the 1st topic entitled Mechanics - a review.
MamaMia!
Thank you for the welcome and for the suggestion. I did not see the homework forum or I would have posted there initially. (Oops! Sorry!)

I will be sure to do so in the future, but feel sure I have been spoiled here by RPenner's TLC.

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