How Fast Is the Skier Going Before She Lands?

[Parzival]

Homework Statement

An extreme skier, starting from rest, coasts down a mountain that makes an angle of 25.0° with the horizontal. The coefficient of kinetic friction between her skis/snow is 0.200. She coasts for 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?

Homework Equations

W = (F cos theta)*displacement
W = KEf - KE0

Kinetic friction = coefficient of kinetic friction * normal force

Weight = mg

The Attempt at a Solution

I tried combining all of these, but i need the mass to do something. I simply can't figure out the final velocity with the information given.

 

[Norfonz]

You can look at the problem in three main segments.
1) Initial Point
2) Top of Cliff
3) Landing at Bottom

Remember that energy is conserved. At rest, the person has some potential energy and no kinetic energy. This total energy is equal to the total energy at point 2, and the total energy at point 3. What type of energy does the person have at point 2?

It may help you to sketch the problem.

 

[Parzival]

You can look at the problem in three main segments.
1) Initial Point
2) Top of Cliff
3) Landing at Bottom

Remember that energy is conserved. At rest, the person has some potential energy and no kinetic energy. This total energy is equal to the total energy at point 2, and the total energy at point 3. What type of energy does the person have at point 2?

It may help you to sketch the problem.

Please clarify. I've been stumped by this for days...

 

[Doc Al]

I tried combining all of these, but i need the mass to do something.

You'll find that the mass will simply drop out of your equations. You won't need a numerical value. Just call the mass 'm' and continue to set up your equations.

How does the total mechanical energy at the start compare to the total mechanical energy when she reaches the edge of the cliff?

 

[Parzival]

You'll find that the mass will simply drop out of your equations. You won't need a numerical value. Just call the mass 'm' and continue to set up your equations.

How does the total mechanical energy at the start compare to the total mechanical energy when she reaches the edge of the cliff?

I can't use any other methods except the formulae given, unfortunately.

 

[Doc Al]

I can't use any other methods except the formulae given, unfortunately.

No problem. Find the net force on the skier and compute the work done as she coasts.

 

[Parzival]

No problem. Find the net force on the skier and compute the work done as she coasts.

But to do that, I need to find the mass of the skier. Can you show me specifically how to?

 

[Saitama]

But to do that, I need to find the mass of the skier. Can you show me specifically how to?

Can you show us the equations you formed?

 

[Parzival]

Please... I have no clue what to do.

 

[Doc Al]

But to do that, I need to find the mass of the skier.

No you don't. Don't try to find a numerical value, just express it symbolically.

Can you show me specifically how to?

What forces act on the skier?

 

[Parzival]

No you don't. Don't try to find a numerical value, just express it symbolically.

What forces act on the skier?

The normal force, which is equal to mg cos 25° but is balanced out by gravity's y component;
gravity's x component, mg sin 25°;
friction: 0.200 * normal force

 

[Doc Al]

The normal force, which is equal to mg cos 25° but is balanced out by gravity's y component;
gravity's x component, mg sin 25°;
friction: 0.200 * normal force

Excellent!

So what's the net force on the skier? Then you can use W_net = ΔKE to find the speed at which she reaches the cliff edge.

 

[Parzival]

Excellent!

So what's the net force on the skier? Then you can use W_net = ΔKE to find the speed at which she reaches the cliff edge.

So the net force is (mg sin 25°) - friction
= (mg sin 25°) - (0.200 * mg cos 25°)

 

[Doc Al]

So the net force is (mg sin 25°) - friction
= (mg sin 25°) - (0.200 * mg cos 25°)

Good. Now find the work done by that net force.

 

[Parzival]

Good. Now find the work done by that net force.

sin 25° is around 0.423, and cos 25° is around 0.906, so:

F = 0.423mg - 0.200*0.906mg
= 0.423mg - 0.181mg
= 0.242 mg

Using W = (F cos θ)* displacement, this is what I get:

(0.242mg cos 0°)*displacement


= 0.242mg * displacement

But what is the displacement?

 

[Doc Al]

But what is the displacement?

That was given in the problem statement.

 

[Parzival]

That was given in the problem statement.

13.9 m?

 

[Doc Al]

13.9 m?

No. We are only dealing (for now) with the part where she coasts down the hill.

She coasts for a distance of 10.4 m before coming to the edge of a cliff.

 

[Parzival]

No. We are only dealing (for now) with the part where she coasts down the hill.

So 10.4 m?

 

[Doc Al]

So 10.4 m?

Yes.

 

[Parzival]

Yes.

So:
W = 0.242mg * 10.4
= 2.52mg

W = KEf (since KE0 is 0)

2.52mg = 1/2m v^2
5.04g = v^2

5.04 * 9.80 = v^2

49.4 = v^2

v = 7.02 which is still wrong. I think that the displacement should equal 13.9 or something.

 

[Doc Al]

So:
W = 0.242mg * 10.4
= 2.52mg

W = KEf (since KE0 is 0)

2.52mg = 1/2m v^2
5.04g = v^2

5.04 * 9.80 = v^2

49.4 = v^2

v = 7.02 which is still wrong.

Assuming you did your arithmetic correctly, what you have found is her speed when she leaves the edge of the cliff. That's not the final answer. You still have to include the effect of her falling from cliff to the final landing point. Keep going.

 

[Parzival]

Assuming you did your arithmetic correctly, what you have found is her speed when she leaves the edge of the cliff. That's not the final answer. You still have to include the effect of her falling from cliff to the final landing point. Keep going.

So:

v^2 = v0^2 + 2ax

v0 = 7.02
a = 9.80 m/s^2
x = 3.50 m

v = sqrt(49.2 + 68.6)

= sqrt(118)

= 10.9 which is correct!

 

[Doc Al]

Good, but be careful!

So:

v^2 = v0^2 + 2ax

This is true, but the reasoning should be from the W = ΔKE theorem, not from kinematics. That result is usually taught as a kinematic relationship for straight line motion--which this is not. The skier leaves the cliff at an angle.

Here's how you would get the same result using the work-energy theorem. Once she leaves the slope, she's in the air. The only force on her is gravity. And the work done by gravity as she falls is mgx, thus:
W = ΔKE
mgx = 1/2mv^2 - 1/2mv0^2
v^2 = v0^2 + 2gx