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.)
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? (Give the magnitude of the acceleration.)

Explanation / Answer

The maximum force of friction = 0.280(38)(9.8) = 104.27 N

Because of the truck's acceleration an implied force of 38 a will be applied to the crate

38a = 104.27
a = 2.74 m/s^2 is the maximum acceleration before the crate moves.

Now once the crate begins to move the friction force drops to 0.16(38)(9.8) = 59.58N

The implied force = 104.27from the first part

Net force = 104.27 - 59.58 =44.68 N

38a = 44.68

a = 1.175 backwards in relation to the truck, but because the truck is accelerating at 2.74m/s^2 forward

the acceleration of the crate in relation to the ground = 2.74 - 1.175 = 1.564 m/s^2 forward.

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