Physics, asked by mirancbsc, 6 months ago

A body of mass 5 kg is kept on rough horizontal surface as shown in figure. The work done by frictional force in time interval t = 0 to t = 3s is (g = 10 ms–2)

Answers

Answered by duttasuman774
2

Explanation:

Normal Reaction on the body=N=mg=5×10=50N

Horizontal force acting on the bodyF

H

=25N

Frictional force acting on the body F

f

=μN=0.2×50=10N

Net force acting on the body F

net

=F

H

−F

f

=25−10=15N

By work-energy theorem,

KE acquired by the body= Work done on the body=F

net

.s=15×10=150J

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