Laws of Motion

JEE Main level, single correct, medium

146 of 210

Question 146 of 210

A body of mass 10kg10kg is moving with an initial speed of 20ms−120m{s}^{-1}. The body stops after 5s5s due to friction between body and the floor. The value of the coefficient of friction is: (Take acceleration due to gravity g=10ms−2g=10m{s}^{-2})

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Answer: (D) 0.40.4

Friction force wil act opposite to the motion of the body.
F=μN=μmgF=\mu N=\mu mg
⇒a=−μg(-ve sign for direction relative to the motion of body)\Rightarrow a=-\mu g\left(\text{-ve sign for direction relative to the motion of body}\right)
Now, applying equation of motion, we get
v=u+atv=u+at
⇒0=20+(−μ×10)×5\Rightarrow 0=20+\left(-\mu \times 10\right)\times 5
⇒50μ=20\Rightarrow 50\mu =20
⇒μ=0.4\Rightarrow \mu =0.4

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