A skier with a mass of 56kg starts from rest and skies down an icy slope that ha

Question

A skier with a mass of 56kg starts from rest and skies down an icy slope that has a length of 63 m at an angle of 32 with respect to the horizontal. At the bottom of the slope, the path levels out and becomes horizontal, the snow becomes less icy, and the skier begins to slow down, coming to a rest in a distance of 120 m along the horizontal path.

A. What is the speed of the skier at the bottom of the slope in m/s?

I have Vf = 2(5.2 m/s)(63 m).

It comes out to 655.2 but the actual answer should be closer to 30 -35 m/s. How do I get from 655.2 to the final answer????

Explanation / Answer

Let coeff of friction be

Then acceleration while sliding down = g(sin-cos)

v^2-u^2 = 2as

v = (2g(sin-cos)*63)

For the horizontal case, acceleration = -g

So, applying v^2-u^2 = 2as again,

v = (2g*120)

Equating both v's we get = 0.1925

From this we get v = 21.278 m/s

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