IP A flatbed truck slowly tilts its bed upward to dispose of a crate. For small

Question

IP A flatbed truck slowly tilts its bed upward to dispose of a crate. For small angles of tilt the crate stays put, but when the tilt angle exceeds 17.5 , the crate begins to slide. When the crate reaches the bottom of the flatbed, after sliding a distance of 2.65 m , its speed is 3.02 m/s .

Find the coefficient of static friction

Find the coefficient of kinetic friction between the crate and the flatbed

I kind of get it, but don't want to risk my last chance on a possibly wrong answer, so thanks a ton in advance!

Explanation / Answer

coeff static friction
µs m g cos 17.5 = mg sin 17.5

µs = sin 17.5 /cos 17.5 = tan0 17.5 = 0.3153

Kinetic friction
Average velocity = ( 0 + 3.02) /2 = 1.51 m/s
time for 2.65 m

t = e / 1.51 = 2.65 /1.51 = 1.755 s
e = 1/2 a * t^2

a = 2e / t^2

a = 2 * 2.65 / 1.755 ^2 = 1.721 m/s2
a = g sin 17.5 - µk g cos 17.5 = g ( sin 17.5 - µk cos 17.5)
a / g = sin 17.5 - µk cos 17.5

µk =0.1312

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