A flatbed truck carries a load of steel. Only friction keeps the steel from slid

Question

A flatbed truck carries a load of steel.  Only friction keeps the steel from sliding on the bed of the truck.  The driver finds that the minimum stopping distance in which he can stop from a speed of 20.0 mi/h without having the steel slide forward into the truck cab is 20.0 m.  What would his minimum stopping distance be when going down a 10 degree incline?

Explanation / Answer

u = 20 mi/h = 20 * 1609 / 3600 = 8.938 m/s s = 20.0 m The acceleration of the truck is                 v2 - u2 = 2as The final velocity is zero. so that we have                 - u2 = 2as            ==> a = - u2 / 2s                       = - (8.938)2 / 2 * 20                       = - 1.997 m/s2 The frictional force between truck and steel is          f = mg Therefore the force along the direction of accelerationis       F = f ==>   ma = mg ==> = a /g = 1.997 / 9.8 = 0.2038 The acceleration of the truck on the inclined plane is           a = g (sin - cos)             = 9.8 * (sin10 - 0.2038 * cos10)             = - 0.2651 m/s2 The stopping distance is            s = -u2 / 2a             = - (8.938)2 / 2 * -0.2651            = 150.64 m

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