A crate of mass 45.0 kg is being transported on the flatbed of apickup truck. Th

Question

A crate of mass 45.0 kg is being transported on the flatbed of apickup truck. The coefficient of static friction between the crateand the truck's flatbed is 0.370, andthe coefficient of kinetic friction is 0.230. (a) The truck accelerates forward on levelground. What is the maximum acceleration the truck can have so thatthe crate does not slide relative to the truck's flatbed?
1 m/s2

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

(a) The truck accelerates forward on levelground. What is the maximum acceleration the truck can have so thatthe crate does not slide relative to the truck's flatbed?
1 m/s2

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

Explanation / Answer

(a) max acc of truck = max acc of box due tostatic friction = force / mass = u m g /m = .           = u g = 0.370 * 9.80 =    3.626 m/s2 . (b) the only force acting on the crate horizontally isthe force of kinetic friction, so its acc relative to the groundis... .            acc = force / mass = u m g / m = u g = 0.230 * 9.80 =   2.254 m/s2 . (b) the only force acting on the crate horizontally isthe force of kinetic friction, so its acc relative to the groundis... .            acc = force / mass = u m g / m = u g = 0.230 * 9.80 =   2.254 m/s2

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