Question

A 70.0kg skiier slides down a hill from a height of 125m. His initial speed is 5m/s.

a) If there is no friction or air resistance, what is his speed when he is at a height of 80.0m? ??????????????

b) If the coefficient of friction between him and the snow is 0.100, and the hill has an angle of elevation of 3??0°, what is his speed when he reaches 80.0m?(neglect air resistance still) ????????????

c) Now (at a height of 80m), the skier moves onto flat ground with a different coefficient of friction. It takes him 14s from the time he reaches the flat surface to slow to a stop. What was the coefficient of friction?

Answer 1

here,

initial speed , u = 5 m/s

initial height , h0 = 125 m

a)

let the final speed be v

using conservation of energy

0.5 * m * u^2 + m * g * h0 = 0.5 * m * v^2 + m * g * h

0.5 * 5^2 + 9.81 * 125 = 0.5 * v^2 + 9.81 * 80

solving for v

v= 30.1 m/s

b)

uk = 0.1

theta = 30 degree

let the final speed be v

using work energy theorm

work done by friction = change in total energy

uk * m * g * cos(theta) * (h0 - h) /sin(theta) = (0.5 * m * u^2 + m * g * h0) - 0.5 * m * v^2 + m * g * h

0.1 * 9.81 * cos(30) * ( 125 -80)/sin(30) = 0.5 * 5^2 + 9.81 * 125 - (0.5 * v^2 + 9.81 * 80)

solving for v

v = 27.5 m/s

c)

let the coefficient of frcition be uk'

accelration , a = uk' * g

v = u + a * t

0 = 27.5 - uk' * 9.81 * 14

solving for uk'

uk' = 0.2