JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 10)
A body of mass $10 \mathrm{~kg}$ is moving with an initial speed of $20 \mathrm{~m} / \mathrm{s}$. The body stops after $5 \mathrm{~s}$ due to friction between body and the floor. The value of the coefficient of friction is:
(Take acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$ )
(Take acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$ )
0.3
0.2
0.5
0.4
Explanation
$a=-\mu g$
$$ \begin{aligned} & \because v=u+a t \\\\ & 0=20+(-\mu \times 10) \times 5 \\\\ & 50 \mu=20 \\\\ & \mu=\frac{2}{5}=0.4 \end{aligned} $$
$$ \begin{aligned} & \because v=u+a t \\\\ & 0=20+(-\mu \times 10) \times 5 \\\\ & 50 \mu=20 \\\\ & \mu=\frac{2}{5}=0.4 \end{aligned} $$
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