Frictional force
Classical Mechanics
Level
3
A block slides down a rough incline sloped at an angle of 40.0 from the horizontal. Starting from rest, it slides a distance of 0.800m down the slope in 0.600 s. What is the coefficient of kinetic friction for the block and surface?(assume g=9.8 m/s^2)
The answer is 0.247.
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First lets find the blocks acceleration; with the x axis pointed down the slope x = 0 + 1/2at^2 =)- a =2x/t^2 Plug in the numbers and get ax: a =2(0.800m)/(0.600 s)^2 = 4.44 ms2 We’ve solved the general problem of a block sliding down a rough inclined plane; ax = g sinΘ − μkg cos Θ = g(sinΘ − μk cosΘ ) where is the angle of the incline. Since μk is the only thing we don’t know here, we do sin 40 − μk cos 40 =axg=(4.44 ms2 )(9.80 ms2 )= 0.454 μk cos 40 = sin 40 − 0.454 = 0.189 μk =(0.189)/cos 40 = 0.247 So we get a coefficient of friction of 0.247 for the block sliding on the surface.