A 16 kg block is dragged over a rough, horizontal surface by a constant force of 188 N acting at an angle of 33.2 ◦ above the horizontal. The block is displaced 96.1 m, and the coefficient of kinetic friction is 0.147. Find the magnitude of the work done by the force of friction.

Explanation:

Given

mass of block

Force

Force is applied at an angle of

Displacement of block

coefficient of kinetic friction

Friction force acting on block

where N=Normal reaction

Work done by friction

Negative sign indicates that displacement is against the direction of force.

Magnitude of work done W = 2.214 kW.

Explanation:

Given :

Mass of object , m = 16 kg.

Angle from horizon , .

Force applied , F = 188 N.

Displacement of block , D = 96.1 m.

Coefficient of kinetic friction , .

Component of force, required to move the block,

Also , frictional force ,

We know, work done ,   ( Here , because the direction of displacement is opposite the direction of frictional force .)

Putting all value in above equation :

We get ,

Magnitude of work done W = 2.214 kW.

Hence, this is the required solution.

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