Calculate Final Speed of 58kg Skier on 25° Slope w/ 70N Friction

[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.

 

[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

 

[Flinze]

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

We have not learned how to use PE yet

 

[rpthomps]

We have not learned how to use PE yet

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.

 

[Flinze]

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.

Thank you! I got it!