A 150.0-kg crate rests in the bed of a truck that slows from 50.0 km/h to a stop in 12.0 s. The coefficient of static friction between the crate and the truck bed is 0.645. What is the minimum stopping time for the truck in order to prevent the crate from sliding?

By Newton's second law,

• the net force acting vertically on the crate is 0, and

∑ F = n - mg = 0   ==>   n = mg = 1470 N

where n is the magnitude of the normal force; and

• the net force acting in the horizontal direction on the crate is also 0, with

∑ F = f - b = 0   ==>   b = f = µn = 0.645 (1470 N) = 948.15 N

where b is the magnitude of the braking force, f is (the maximum) static friction, and µ is the coefficient of static friction. This is to say that static friction has a maximum magnitude of 948.15 N. If the brakes apply a larger force than this, then the crate will begin to slide.

Note that we are taking the direction of the truck's motion as it slows down to be the positive horizontal direction. The brakes apply a force in the negative direction to slow down the truck-crate system, and static friction keeps the crate from sliding off the truck bed so that the frictional force points in the positive direction.

Let a be the acceleration felt by the crate due to either the brakes or friction. Use Newton's second law again to solve for a :

f = ma   ==>   a = (948.15 N) / (150.0 kg) = 6.321 m/s²

With this acceleration, the truck will come to a stop after time t such that

0 = 50.0 km/h - (6.321 m/s²) t   ==>   t ≈ (13.9 m/s) / (6.321 m/s²) ≈ 2.197 s

and this is the smallest stopping time possible.

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