Question

# Sam takes off from rest down a 50m high 10 degree slope on his jet-powered skis....

Sam takes off from rest down a 50m high 10 degree slope on his jet-powered skis. The jets on his skis still have a thrust of 200 N and Sam's mass is still 75kg.

If sam's speed is 40m/s at the bottom of the hill, what is the coefficient of kinetic friction between his skis and the snow?

here,

mass of Sam , m = 75 kg

height , h = 50 m

theta = 10 degree

thrust due to jets , Ft = 200 N

normal force , N = m * g * cos(theta)

the length of slope ,s = h/sin(theta)

let the coefficient of kinetic friction be uk

using work energy theorm

total work done = kinetic energy gained

Wt + Wg + Wff = KE

Ft * s + m * g * H - uk * N * s = 0.5 * m * v^2

Ft * (h /sin(theta)) + m * g * H - uk * N * (h/sin(theta)) = 0.5 * m * v^2

200 * ( 50/sin(10)) + 75 * 9.81 * 50 - uk * 75 * 9.81 * cos(10) * ( 50/sin(10)) = 0.5 * 75 * 40^2

solving for uk

uk = 0.16

the coefficient of kientic friction is 0.16

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