A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. T

Question

A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck's flatbed is 0.330, and the coefficient of kinetic friction is 0.220.
(a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck's flatbed?
__________ m/s2

(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?
_________ m/s2

Explanation / Answer

(a)

Us = coefficient of static friction
Fs = static friction = (normal force)*Us = m*g*Us

Force pulling the crate = m*a
m*a = m*g*Us
a = g*Us = 9.81*0.27 = 2.64 m/s^2

(b)
Uk = coefficient of kinetic friction
Fk = kinetic friction = m*g*Uk
g*Uk = acceleration due to friction

net acceleration = g*Us - g*Uk = g(Us - Uk)
a = 9.81(0.27 - 0.24) = 9.81*0.129
a = 0.2943 m/s^2

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