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 Fk= μkmg = 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.
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