# A 21.6 kg box initially at rest is pushed 7.88 m along a rough, horizontal floor with a constant applied horizontal force of 86.9829 N. If the coefficient of friction between box and floor is...

A 21.6 kg box initially at rest is pushed 7.88 m along a rough, horizontal floor with a constant applied horizontal force of 86.9829 N. If the coefficient of friction between box and floor is 0.341, find the work done by the friction and the final speed of the box.

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### 1 Answer

A 21.6 kg box initially at rest is pushed 7.88 m along a rough, horizontal floor with a constant applied horizontal force of 86.9829 N. The coefficient of friction between box and floor is 0.341.

The frictional force acting on the box is F = N*C, where N is the normal force and C is the coefficient of friction.

F = 21.6*9.8*0.341 = 72.18 N. The work done by the force as the box is pushed along a path 7.88 m long is 568.8 J

The net force acting on box is the frictional force subtracted from the applied force. This is equal to : 86.9829 - 72.18 = 14.80 N. The force results in an acceleration of the box of 14.80/21.6 m/s^2.

As the box starts from rest, the final speed is v given by v = sqrt(2*14.80*7.88/21.6) = 3.286 m/s

The final speed of the box is 3.286 m/s and the work done by the force of friction is 568.8 J