A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at

Question

A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at an angle ? from the horizontal. The block is originally located 5m from the bottom of the plane.

If the block, undergoing constant acceleration down the ramp, slides to the bottom in t = 2 s, and ? = 30

A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at an angle ? from the horizontal. The block is originally located 5m from the bottom of the plane. If the block, undergoing constant acceleration down the ramp, slides to the bottom in t = 2 s, and ? = 30 degree, what is the magnitude of the kinetic frictional force on the block?

Explanation / Answer

a = g - mue*g*sin(30) --(1)

s = 0.5*a*t^2

==> a = 2*s/t^2 = 2*5/2^2 = 2.5 m/s^2

from equation 1

2.5 = 9.8 - mue*9.8*sin(30)

==> mue = (9.8-2.5)/(9.8*sin(30)) = 1.4898

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