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$300 \mathrm{~J}$ of work is done in sliding a $2 \mathrm{~kg}$ block up an inclined plane of height 10 $\mathrm{m}$. Taking $g=10 \mathrm{~m} / \mathrm{s}^2$, work done against friction is:
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Verified Answer
The correct answer is:
$100 \mathrm{~J}$
Potential energy
$$
\begin{aligned}
& =2 \times 10 \times 10 \\
& =200 \mathrm{~J}
\end{aligned}
$$
and work done $=300 \mathrm{~J}$
$\therefore$ Work done against friction
$$
=300-200=100 \mathrm{~J} .
$$
$$
\begin{aligned}
& =2 \times 10 \times 10 \\
& =200 \mathrm{~J}
\end{aligned}
$$
and work done $=300 \mathrm{~J}$
$\therefore$ Work done against friction
$$
=300-200=100 \mathrm{~J} .
$$
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