Question #116744
I drag a mass m=21 kg in a straight line, along a horizontal surface, a distance D= 23m .I drag it at constant speed v=0.90m/s in a straight line using a horizontal force.the coefficient of static friction is 1.2 and coefficient of kinetic friction is 1.1.How much work do I do in joules?
1
Expert's answer
2020-05-18T10:22:00-0400

As the object is in motion hence kinetic friction will take place

kinetic friction = μk×N=1.1×21×10\mu_k\times N=1.1\times21\times10 =231N231 N

lets apply work energy theorem

W=ΔK\sum W=\Delta K

WN+Wf+Wmg+WF=ΔKW_N+W_f+W_{mg}+W_F=\Delta K

0(231×23)+0+WF=12×21×0.920-(231\times23)+0+W_F=\frac12\times21\times0.9^2

WF=5321.5JW_F=5321.5 J




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