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A force of $20 \mathrm{~N}$ is applied on a body of mass $5 \mathrm{~kg}$ resting on a horizontal plane. The body gains a kinetic energy of $10 \mathrm{~J}$ after it moves a distance $2 \mathrm{~m}$. The frictional force is
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Verified Answer
The correct answer is:
$15 \mathrm{~N}$
Given, force $F=20 \mathrm{~N}$
$$
m=5 \mathrm{~kg}
$$
Net force on the body,
$$
F_{\text {net }}=F-f_{s} \quad \text { [where, } \int_{s}=\text { friction] }
$$
$\therefore$ Work done by displacing $2 \mathrm{~m}$ is given as
$$
\begin{aligned}
W &=F_{\text {net }} \times \text { displacement } \\
&=\left(F-f_{s}\right) 2 \\
&=\left(20-f_{s}\right) 2 \\
\Rightarrow \quad W &=40-2 f_{s}
\end{aligned}
$$
According to work-energy theorem,
Work done = gain in kinetic energy
$$
\begin{aligned}
\Rightarrow & 40-2 f_{s} &=10 \\
\Rightarrow & 2 f_{s}=40-10 &=30 \\
\Rightarrow & f_{s} &=\frac{30}{2} \\
&=15 \mathrm{~N}
\end{aligned}
$$
$$
m=5 \mathrm{~kg}
$$
Net force on the body,
$$
F_{\text {net }}=F-f_{s} \quad \text { [where, } \int_{s}=\text { friction] }
$$
$\therefore$ Work done by displacing $2 \mathrm{~m}$ is given as
$$
\begin{aligned}
W &=F_{\text {net }} \times \text { displacement } \\
&=\left(F-f_{s}\right) 2 \\
&=\left(20-f_{s}\right) 2 \\
\Rightarrow \quad W &=40-2 f_{s}
\end{aligned}
$$
According to work-energy theorem,
Work done = gain in kinetic energy
$$
\begin{aligned}
\Rightarrow & 40-2 f_{s} &=10 \\
\Rightarrow & 2 f_{s}=40-10 &=30 \\
\Rightarrow & f_{s} &=\frac{30}{2} \\
&=15 \mathrm{~N}
\end{aligned}
$$
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