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 = 22 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 to avoid your answer being counted wrong due to rounding. You answer will not be graded on the number of significant figures you give.

Explanation / Answer

FBD of body
mgsin(theta) - f = ma
and f = muN
anf N = mgcos(theta)
so f = mu*mgcos(theta)
and mgsin(theta) - mu*mgcos(theta) = ma
a = g(sin(theta) - mu*cos(theta)) = 9.8(0.4226 - mu*0.9063) = 4.1414 - mu*8.8811-- (1)

v at bottom of plank
v = sqroot(2*a*s) -- (2)

Now deceleration on horizontal track = mu*9.8
and 2*mu*9.8*22 = 2*[4.1414 - mu*8.8811]*12
mu = 0.1542

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