A 60.2 kg skier is sitting In a 139.8 kg toboggan cart. A machine pushes the fob

Question

A 60.2 kg skier is sitting In a 139.8 kg toboggan cart. A machine pushes the foboggan cart off the top of a frictionless bill with an initial speed of 4.00 m/s at point A. The cart slides down a frictionless ramp which becomes horizontal (at point C) after dropping 10.0m. The cart experiences no significant air resistance. Using work energy principles, answer the following: At what height (point B) above the horizontal section is the cart moving at 8.00 m/s? Drawing is approximate. The ice on the horizontal section is very smooth till point D and then it becomes rought. Son the toboggan cart with skier stops after travelling a distance d of 63.6m (between D and G) on the rought patch of ice, (i) What is the speed of the cart between C and D and (ii) how much is the coefficient of kinetic friction of ice in the rough patch?

Explanation / Answer

(a) Work energy theoram

W = change in KE

F.s = KE at height h - KE at 10m height (here h=s)

mg x s = 1/2 m (8)^2 - 1/2 m (4)^2

9.8 x s = 1/2 (64 - 16) = 24

s = 2.44 m

means from ground = 10 - 2.44 = 7.56 m

(b) (1) Work-Energy theroram

F.s = KE at D -KE at height 10 m

mg x s = 1/2 m (v)^2 - 1/2 m (4)^2

9.8 x 10 = 1/2 x (v^2 - 16)

v = 14.56 m/s

(2) Again work-energy theram

W = KE at D - KE after d distance

f.d = 1/2 m v^2 - 0

f x 63.6 = 1/2 x 200 x (14.56)^2

f = 333.32 N

f = mu x N = mu x mg

333.32 = mu x 200 x 9.8

mu = 333.32 / (200 x 9.8)

mu = .17

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