A 69.8 kg snowboarder starts from rest at the top of a hill of height H = 17.9 m

Question

A 69.8 kg snowboarder starts from rest at the top of a hill of height H = 17.9 m as shown in the diagram. The hill is inclined at an angle of 22 degrees relative to the horizontal. The hill is frictionless, but the horizontal surface at the bottom of the hill is rough. The coefficient of kinetic friction between the snowboard and the horizontal surface is 0.17.

What is the snowboarder's speed when he reaches the bottom of the hill?

How far does the snowboarder travel (d) after reaching the bottom of the hill before he comes to rest?

Please show all work, and answer in sig-figures

Hint** Make sure you use theta = 180 and not what is given in the problem, that is what was giving me the wrong answer.

Thank you for all your help

Explanation / Answer

a)so using energy let the speed be V

1/2 mV2 = mgH

V = 18.73 m/s

b) let it be D

V2 = -2aD

a = -g = -0.17*9.8 = -1.666

D = 105.3 meters

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