Class Management Help Hw 6 Begin Date: 2/16/2018 12:00:00 AM - Due Date: 2/28/20
ID: 1581202 • Letter: C
Question
Class Management Help Hw 6 Begin Date: 2/16/2018 12:00:00 AM - Due Date: 2/28/2018 11:59:00 PM End Date: 5/31/2018 12:00:00 AM (8%) Problem 12: A skateboarder with mass ms-35 kg is standing at the top of a ramp which is hy-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 vf= 6.1 m/s 33% Part (a) 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 33% Part (b) The ramp makes an angle with the ground, where = 30°. Write an expression for the magnitude of the friction force, , bctwccn thc ramp and thc skatcboardcr. > 33% Part (c) 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, Fgrass in newtons, between the skateboarder and the grass Grade Summary Deductions Potential 3% 97% rass sinO cotan)a cosO asin atan)acotan)sinhO cosh0 t cotanhO tan() ( acos() Submissions Attempts remaining: 9 (390 per attempt) detailed view 3% 0 Degrees Radians Submit Hint I give up! Hints: % deduction per hint. Hints remaining: 4 Feedback: 0% deduction per feedback -Explanation / Answer
a) PE = m g h
KE = m v2 / 2
From conservation of energy
Wf + m g h = m v2 / 2
Work done by friction Wf = [m v2 / 2] - m g h
b) Wf = -F d
F = Wf * sin theta / h
c) Wgrass = delta E
E = 1/2 m v2
Wf = m v2 / 2 = Ff * d
Ff = m v2 / 2 d = (35 * 6.12) / (2 * 5.00)
magnitude of fruction force = 130 N