A heavy box of mass 20 kg is pulled on a horizontal surface.If the coefficient of kinetic friction between the box and the horizontal surface is 0.25 find the force of friction exerted by the horizontal forces on the box

Dear Pallavi,

The kinetic friction generated due the horizontal forces on the box is given by the formula F_k= μ_kmg = 0.25 * 20 * 9.8 N = 49N

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Adapa Bharath

While in motion only the kinetic friction acts. f=uN.

                                                               f=umg

                                                                =0.25*20*9.8

                                                                =49N

so the friction exerted by the horizontal forces is 49N.