Work and Energy of an extreme skier

[chantalprince]

Hey,

I have a question that I can't seem to work out...can anybody please help me?!

Question:

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25 degrees with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts down a distance of 10.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.50 m below the edge. How fast is she going just before she lands?


I have set the problem up and come to realize that this will take 2 separate analyses: one before she flies off the cliff and one after to get the final velocity of the second analysis, where v final of the 1st analysis will equal v initial of the 2nd.

What has me stumped is the fact that I was given no mass. In order to to find my kinetic energy (KE = 1/2 mv (initial)^2 + W) I need a mass.

Can anybody please help me?! I am a lost cause at the moment. Your help is MUCH appreciated!

Thank you!

 

[Doc Al]

You won't need the actual mass--just call it "m" and keep going. Hint: What's the force of friction?

 

[chantalprince]

So I got to a point where I think the masses cancel... And I am sorry, I don't quite understand what you mean by: what's the force of friction? I think they gave me a number (0.200)...

 

[Doc Al]

They gave you the coefficient of kinetic friction. Use that to find the friction force. (Look it up if you need to.)

 

[chantalprince]

Ok, so I just realized that they gave the the coefficient of kinetic friction. Fk = coeff. x normal force. From what I understand, the normal force cancels out because there is no accelleration in the y direction...I just can't connect how this information helps me with the mass part of it.

 

[Doc Al]

Fk = coeff. x normal force.

Good.

From what I understand, the normal force cancels out because there is no accelleration in the y direction...

Use that fact to figure out the normal force, and then the friction force.

 

[chantalprince]

So normal force = 0, is this correct? If so, then that makes my friction force = 0. I still can't connect that with the mass part of this. The only thing I can think of is that the friction force balances the weight force and so my mg will equal zero too...??

 

[Doc Al]

So normal force = 0, is this correct?

Not at all. Consider all forces acting on the skier in the y-direction (perpendicular to the slope). As you stated earlier, the acceleration in that direction is zero, thus the net force must be zero. Use that fact to figure out the normal force.

 

[chantalprince]

Ok, so Fn = mgcos(25) So I still have m in there. This is my overall equation:

Square root: 2(mgsin25 - coeff. k(mgcos25)) (cos theta) (s)/ m

I don't get it! Do my masses cancel somehow??

I can see how the masses on top might, but what happens to the denominator? (My algebra is obviously lacking...)

 

[Doc Al]

Ok, so Fn = mgcos(25) So I still have m in there.

Excellent.

This is my overall equation:

Square root: 2(mgsin25 - coeff. k(mgcos25)) (cos theta) (s)/ m

OK. Looks like you're solving for the speed at the point where the skier goes over the cliff. No problem. (What's theta?)

I don't get it! Do my masses cancel somehow??

Well, yes. Look at what you wrote.

I can see how the masses on top might, but what happens to the denominator? (My algebra is obviously lacking...)

It's the denominator that does the canceling in the numerator! ab/a = b

FYI: You don't have to do this in two steps. You can compare initial energy to final energy and use that to solve for the final speed in one step.

 

[chantalprince]

Theta is 0 degrees because the force and the net force and the displacement go in the same direction.

I can use: (Sum of forces) x s = 1/2 mv^2 - 1/2 mv (initial) ^2 and that final velocity is the final after she flies off of the cliff? So I use the second displacement here and the final velocity from the first part for initial...is this correct?

And is my sum of forces the same as above, or do I need a new force diagram, in which the only forces acting after she flies off of the cliff is weight?

 

[Doc Al]

Theta is 0 degrees because the force and the net force and the displacement go in the same direction.

Good.

I can use: (Sum of forces) x s = 1/2 mv^2 - 1/2 mv (initial) ^2 and that final velocity is the final after she flies off of the cliff?

Right. This is what I thought you did.

So I use the second displacement here and the final velocity from the first part for initial...is this correct?

Right. Hint: You'll save time by not bothering to solve for the final velocity from the first part, but just stick with the final KE (which is what you really need).

And is my sum of forces the same as above, or do I need a new force diagram, in which the only forces acting after she flies off of the cliff is weight?

The forces are different--there's no normal force, only weight.

 

[chantalprince]

Thank You So Much Doc Al!