Skier coasting down slope with initial speed and friction

[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 don't 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 don't know where to plug #s in or if this is even the right equation?!
Thanks again!

 

[Päällikkö]

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

Can you take it from here?

 

[BriannaUND]

No- that makes me more confused!

 

[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?

 

[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>?

 

[BriannaUND]

Also, I calculated #72 out, and I'm getting 5.0 m/s but I don't understand how the speed is the same?!

 

[quantumdude]

OK, let's back up here.

I calculated 2gy= v2 and plugged in v= sq root (2x 9.8m/s2 x10m)= 14m/s Does this look correct?

No, it is wrong.

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 don't know how to go on from there!

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.

I know that there is an equation v= sq root (vo2 - 2ukmgd) but I don't know where to plug #s in or if this is even the right equation?!

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.

 

[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