A crate of mass 38.0 kg is being transported on the flatbed of a pickup truck. T

Question

A crate of mass 38.0 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 0.280, and the coefficient of kinetic friction is 0.160.

(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? (Give the magnitude of the acceleration.)
2.74 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
  to the ground? (Give the magnitude of the acceleration.)

Explanation / Answer

Given
Mass of crate, m = 38 kg
coefficient of static friction,µs =0.280
coefficient of kinetic friction,µk = 0.160
a)The net force on the truck is
F = µs m g
ma = µs m g
a = 0.280 *9.8 m/s^2
a =2.74 m/s^2
Thus, the acceleration of the truck is 2.74 m/s^2
b) The net force on the crate is
F = µk m g
m a = µk m g
a = 0.160 *9.8 m/s^2
a = 1.568 m/s^2

Thus the acceleration of the crate relative to ground is 2.74-1.568= 1.172 m/s^2---ANSWER b)

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