# Inclined Plane

1. Feb 11, 2008

### 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?

Last edited: Feb 11, 2008
2. Feb 11, 2008

### Staff: Mentor

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.

3. Feb 11, 2008

### 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.

4. Feb 11, 2008

### Staff: Mentor

I assume that this:
means that they want the time it takes for her to come to rest.

5. Feb 12, 2008

### 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?

6. Feb 13, 2008

### Staff: Mentor

Show us exactly what you did.

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

7. Feb 13, 2008

### 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.

8. Feb 13, 2008

### Staff: Mentor

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.