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A body of mass $10 \mathrm{~kg}$ is kept on a rough horizontal surface of coefficient of friction 0.3 . If a horizontal force of $50 \mathrm{~N}$ is applied on the body, then the acceleration of the body is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$2 \mathrm{~ms}^{-2}$
Mass, $m=10 \mathrm{~kg}$
Coefficient of friction, $\mathrm{M}=0.3$
force, $\mathrm{F}=50 \mathrm{~N}$.
Acceleration, $\mathrm{a}=$ ?
Net Force is given by,
$F-f_s=m a$
Here, $\mathrm{f}_{\mathrm{s}}=\mu \mathrm{mg}$ [frictional force]
$\begin{aligned} & 50-\mu \mathrm{mg}=\mathrm{ma} \\ & 50-0.3 \times 10 \times 10=10 \times \mathrm{a} \\ & 10 \mathrm{a}=20 \\ & \mathrm{a}=2 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
Coefficient of friction, $\mathrm{M}=0.3$
force, $\mathrm{F}=50 \mathrm{~N}$.
Acceleration, $\mathrm{a}=$ ?
Net Force is given by,
$F-f_s=m a$
Here, $\mathrm{f}_{\mathrm{s}}=\mu \mathrm{mg}$ [frictional force]
$\begin{aligned} & 50-\mu \mathrm{mg}=\mathrm{ma} \\ & 50-0.3 \times 10 \times 10=10 \times \mathrm{a} \\ & 10 \mathrm{a}=20 \\ & \mathrm{a}=2 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
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