A 77.5-kg crate is sitting on the ground when an external force is applied. The

Question

A 77.5-kg crate is sitting on the ground when an external force is applied. The coefficient of kinetic friction for these surface is 0.12, and the coefficient of static friction for these surfaces is 0.20. For each scenario described below, determine if the friction on the crate will be static or kinetic and calculate the magnitude of the friction. a) A 120-N force is applied to the stationary crate for 10 seconds. After 5 seconds, the friction is .b.) a 180N is applied to the stationary crate for 10sec. After 5sec the friction is

Explanation / Answer

The mass of the crate = 77.5 kg, when it is stationary, The force acting is gravity downwards. hence The Normal force Fn down wards = mg= 77.5 x 9.8 = 759.5 N

The static friction given is = 0.2

The Budging force = FB= 759.5 x 0.2 =151.9 N, hence minimum 151.9N force is required to move the crate from stationary. But the forceapplied is 120N hence the net force acting in the direction of applied force= 151.9-120=31.9N

b) in second case, 180N is applied hence , the object starts moving in forward direction and net Force= 180-0.12*759.5=91.14 N (as coefficient of kinetic friction comes into picture)

the accelaration = F/m = 91.14/77.5=1.176m/s2 if it applied for 5 seconds, the velocity = 1.176 x5 = 5.88 m/s

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