Question

A skateboarder with mass ms = 42 kg is standing at the top of a ramp which is hy = 3.7 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.3 m/s.

Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.

The ramp makes an angle θ with the ground, where θ = 30 degrees. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.

When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, fg in Newtons, between the skateboarder and the grass.

Answer 1

W(fr)=PE-KE = mgh - mv²/2
W(fr) =F(fr) •s=F(fr)•h/sinα
KE=W(fr)₁
mv²/2 =F(fr)₁ •s₁
F(fr)₁= mv²/2•s₁