During the winter break you go sledding with your friends. The hill you are sled

Question

During the winter break you go sledding with your friends. The hill you are sledding on is L = 12 m long (that's the distance along the slope, not the vertical height of the slope) and slopes at an angle of = 25 degrees. At the bottom of the hill is a long horizontal stretch of snow where you slow down and stop.

If you start with zero velocity at the top of the hill and you stop a distance d = 21 m from the bottom of the hill, what is the coefficient of friction, k of the sled on the snow? Assume it is the same on both the hill and along the horizontal.

Give your answer to at least three significant figures.

Explanation / Answer

given theta = 25 degrees,

L = 12 m

so height (H) = L * cos ( theta) = 12* cos(25) = 11.0865 m

now let us use conservation of energy.

potential energy at top = velocity at bottom of hill.

let the velocity be V at bottom

so,

mgH = 0.5mV2

=>V= sqrt( gH) = sqrt (9.81*11.0865)= 10.4287 m/s

now , the sled stops after a distance of 21 m

so work done by friction =- k * m*g * d = change in kinetic energy

so ,

we have

final kinetic energy- initial kinetic energy = work done

=> 0- 0.5mV2= - k * m*g * d

=> 0.5*( 10.4287 m/s)2 = k * 9.81 * 21

=> k= 0.2639

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