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A 5.9 kg box slides a 9.3 m distance on ice. If the coefficient of kinetic friction is 0.206, what is the work done by the friction force

Sagot :

Work done by the friction force is 110.77 J

Calculation of friction force:

here box has mass, m=5.9 kg

It slides the distance of, d=9.3 m

coefficient of kinetic friction is 0.206

Now we can say that,

net force on the object is applied force and friction force

The net force according to Newton's second law:

F(net)=m*a

F(applied) - f(friction) = m*a

The work done by friction force will be

W= -k*mg*d

here g is gravitational acceleration.

substituting the values we get,

W=-(0.206)*(5.3)(9.8)*(9.3)

W= -110.77 J

negative sign denotes the opposite direction.

work done by the friction force is 110.77 J

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