Question

A man pushes a box initially at rest at towards another man by exerting a constant horizontal force of magnitude 5 newton through a distance of 1 metre its final kinetic energy is​

Answer 1

Answer:

by applying the work energy theorem i.e.

W net = Δ K.E.

work done = force applied x displacement

                   =  5 N x 1 m

                   =  5 J

now ATQ initial velocity is 0 therefore initial kinetic energy will be 0 ( K.E. = 0)

and let say final velocity will be K.E. = y

then from work energy theorem

5 J = K.E.f -  K.E.i

5 J = K.E.f

5J = Y

hope this helps mate

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