Laws of Motion

JEE Main level, numerical answer, medium

161 of 210

Question 161 of 210

A boy of mass 4kg4kg is standing on a piece of wood having mass 5kg5kg. If the coefficient of friction between the wood and the floor is 0.5,0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is _______ NN. (Round off to the Nearest Integer)
[Take
g=10ms−2]\left.g=10{ms}^{-2}\right]

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Answer: 30


N+T=90N+T=90
T=μN=0.5(90−T)T=\mu N=0.5(90-T)
1.5T=451.5T=45
T=30NT=30N

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