A boy of mass 40 kg is climbing at a constant of fricrion between his palms and the pole is `0.8 and g=10^-2`, the horizontal force that he is applying on the pole is

Options

A. 300 N
B. 400 N
C. 500 N
D. 600 N

Answer

C

Explanation

As the boy is climbing the pole at a constant the pole at a constant speed(no acceleration),the force of friction must be just balance by his weight, i.e.,`mu R=mg or R=(mg)/mu=(40 xx 10)/0.8 =500N`