A skateboarder with mass m_s = 52 kg is standing at the top of a ramp which is h
ID: 1518306 • Letter: A
Question
A skateboarder with mass m_s = 52 kg is standing at the top of a ramp which is h_y = 3.6 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 v_f = 6.8 m/s. m_s = 52 kg h_y = 3.6 m v_f = 6.8 m/s Write an expression for the work, w_f, 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 theta with the ground, where theta = 30 degrees. Write an expression for the magnitude of the friction force f_r, between the ramp and the skateboarder. When the skateboarder reaches the bottom of the ramp, be continues moving with the speed v_f 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. F_grass in newtons, between the skateboarder and the grass.Explanation / Answer
Here,
m = 52 Kg
hf = 3.6 m
vf = 6.8 m/s
a) Now , for the work done due to frictional force
work done by to frictional force = ms * g * hf - 0.5 * ms* vf^2
work done by to frictional force = ms * (g * hf - 0.5 * vf^2)
b)
let the frictional force is f
f * distance = ms * (g * hf - 0.5 * vf^2)
f * hf/(sin(30) = ms * (g * hf - 0.5 * vf^2)
f = ms * (g * hf - 0.5 * vf^2) * sin(30)/hf
c)
let the frictional force is f
f * 5 = decrease in kinetic energy
f * 5 = 0.5 * 52 * 6.8^2
f = 240.5 N
the magnitude of frictional force acting is 240.5 N