Time for Skier to Stop in 8.7° Slope: No Friction

[Ryo124]

A skier with mass 65.9 kg is going 14.9 m/s on a flat surface. She hits a slope with angle 8.7 degrees from the horizontal. How long does it take for her to stop? Neglect friction.

I've tried breaking up her velocity into components. Also, found the force on her once she is on the slope with mgsin\theta. I don't know if I needed to do this or what to do once I found these values.

I'm just not sure how to start this and what to do from there?

 

[Doc Al]

She starts up the slope with her full speed, so no components needed. You found the force on her, so what's her acceleration? The rest is kinematics.

 

[lavalamp]

Personally I'd just use conservation of energy.

Find her kinetic energy (KE = 0.5 * mv^{2}), find at what height h she would have the same amount of gravitational potential energy (GPE = mgh).

Once you know how high she has to go, and at what angle she has to do it at, it's trigonometry.

 

[Doc Al]

I assume that this:

How long does it take for her to stop?

means that they want the time it takes for her to come to rest.

 

[Ryo124]

I keep getting 5.02 sec. That's not right.

The answer is 10.0 sec. Can anyone walk me through this to show why this answer is correct?

 

[Doc Al]

Show us exactly what you did.

What was the acceleration? What kinematic equation did you use to find the time?

 

[Ryo124]

Never mind, I solved it.

I solved for her force parallel to the plane, F||, and set that equal to ma. Then, I solved for a.

Then, I used the equation: V=Vo + at to solve for t.

However, when I solve for t in the above equation, I don't get why her initial velocity, Vo, is 14.9 m/s and not her y-component of velocity, which would be about 2 m/s.

 

[Doc Al]

When she hits the slope we assume no energy is lost. The slope just changes her direction, not her speed--she's still going full speed when she starts up the slope.