Question #104982

I drag a mass m=21 kg in a straight line, along a horizontal surface, a distance D=23 m . I drag it at constant speed v=0.90 m.s−1 in a straight line using a horizontal force. The coefficients of friction are \mu_{s} = 1.2μ

s

​ =1.2 and \mu_{k} = 1.1μ

k

​ =1.1. How much work do I do?

Expert's answer

As the object is in motion then kinetic friction force will act on it and to keep the object in motion we have to apply a constant force which will be equal to the magnitude of kinetic friction force

Normal Force = m×g=21×10=210m\times g=21\times 10=210


fk=μk×Normalfk=1.1×210=21×11=231Nf_k=\mu_k\times Normal\\f_k=1.1\times 210=21\times11=231 N


Work done=Force×displacementWork done=231×23=5313J\\Work \ done=Force\times displacement \\Work \ done=231\times23=5313J


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