Suppose a skateboarder reaches a height of 1.8 m above the right side of a semic

Question

Suppose a skateboarder reaches a height of 1.8 m above the right side of a semicircular ramp. He then reverses himself and reaches the bottom of the ramp with a speed of 6.4 m/s. The skateboarder mass is 64 kg and the radius of the semicircular ramp is 2.70 m.

a) What is the mechanical energy of the skateboarder at before his reverses himself?

b) What is the mechanical energy of the skateboarder when he reaches the bottom of the semicircular ramp?

c) What is the average frictional force exerted by the ramp on the skateboarder?

Explanation / Answer

total height = 1.8 + 2.70 = 4.50 m

GPE at top = m g h = 64 *9.81* 4.50 = 2825.28 J

KE at bottom = 1/2 m v^2 = 1/2 *(64)*(6.40)^2 = 1310.72 J

energy lost to friction = GPE top - KE bottom = 2825.28 - 1310.72 = 1514.56 J  

work done against friction = F*d

and d = 1/4 circumference = 1/4 (2*pi*r) = 1/2 pi (2.70) = 4.2428 m  

1514.56 = F (4.2428)

F = 356.97 N or average 357 N

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