# Work in 2D motion

1. Jan 8, 2015

### Flinze

A skier 58.0kg coasts down a 25.0 degree uniform slope. A kinetic frictional force of 70.0N opposes her motion. At the top of the slope her speed is 3.60 m/s. What is her final speed at a distance 57.0 metres down the slope?

Kinetic Energy = 1/2mv^2
V=Vi+at^2
F=ma

So far I decomposed the force of gravity portion, Fx = (58)(9.8)(cos25) and then I subtrated 70.0 N from that getting 445.145. At this point, I don't know what to do.

2. Jan 8, 2015

### rpthomps

The best approach to this question, I think, is to use conservation of energy. KE (at top) + PE (at top) = KE (at new position) + PE(at new position) + Work done by friction

3. Jan 8, 2015

### Flinze

We have not learned how to use PE yet

4. Jan 8, 2015

### rpthomps

In that case there is an error in your original equation. The force of gravity parallel to the ramp is $$mgsin\theta$$

Subtract the force of friction like you did and then use newton's second law to find acceleration. Then find an appropriate kinematics relation that involves a, two velocities an displacement.

5. Jan 8, 2015

### Flinze

Thank you! I got it!