Homework Help: Homework help

1. Oct 4, 2005

BriannaUND

I'm having problems with the following two problems. Thank you for any help!
72) A skiier coasts down a very smooth 10-meter high slope. If the speed of the skiier on the top of the slope is 5.0m/s, what is his speed at the bottom of the slope?
I calculated 2gy= v2 and plugged in v= sq root (2x 9.8m/s2 x10m)= 14m/s Does this look correct? We did an example like this in class but we used vinitial = 0 m/s where its 5.0m/s in this problem so I dont know how to go on from there!
79) In #72, if the skiier has a mass of 60kg and the force of friction retards his motion by doing 2500 J of work, what is his speed at the bottom of the slope?
I know that there is an equation v= sq root (vo2 - 2ukmgd) but I dont know where to plug #s in or if this is even the right equation?!
Thanks again!

2. Oct 4, 2005

Päällikkö

The principle of conservation of energy:
$$K_A + U_A \pm W = K_B + U_B$$

Can you take it from here?

3. Oct 4, 2005

BriannaUND

No- that makes me more confused!

4. Oct 4, 2005

Päällikkö

Oh, sorry :).
Well the above equation translates in this case into:
$$\frac{1}{2}mv_A^2 + mgh_A \pm W = \frac{1}{2}mv_B^2 + mgh_B$$

Does this help at all?

5. Oct 4, 2005

BriannaUND

ok- i can figure out # 72 now but How do I apply the "force of friction retards his motion by doing 2500 J of work" in #79>?

6. Oct 4, 2005

BriannaUND

Also, I calculated #72 out, and I'm getting 5.0 m/s but I dont understand how the speed is the same?!

7. Oct 4, 2005

Tom Mattson

Staff Emeritus
OK, let's back up here.

No, it is wrong.

Write down an expression for the total energy at the top of the slope, and another expression for the total energy at the bottom. Then apply the law of conservation of energy, just like Päällikkö said. W represents the work done by nonconservative forces. Since #72 doesn't mention friction, you can set W=0 for that one.

It is not the correct equation. Again, you have to apply the law of conservation of energy, and this time the work W is not zero.

Brianna, you seem to be looking for a formula that will solve every problem, instead of thinking about the problem. You can't do physics formulaically. You have to do some analysis wherein you apply physical principles, and in this case the principle is the law of conservation of energy.

8. Oct 4, 2005

BriannaUND

Ok- i'm sorry to keep asking about the same questions but I tried to apply the law of conservation of energy to the two problems and I'm not coming out with the right answers:
72) I understand that PE at the top of the hill will be zero because the object is already moving, and it will also be 0 at the bottom of the hill because y=0. Applying this, that should leave KEa = KEb: .5mv2a= .5mv2b. This gives me .5(5)2 = .5v2 because m will cancel out. After calculating, I'm winding up with 5 m/s but that can't be the correct answer because the speed should increase as the skiier slides down the hill.
79) Again, PE should be zero at the top and bottom of the hill but now W is involved. SO I plugged in .5(60kg)(5)2 +2500 J = .5(60kg)(v)2. I calculated this out and got 10 m/s but I was told the answer is 12 m/s