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A mass of $M \mathrm{~kg}$ is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of $45^{\circ}$ with the initial vertical direction is
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Verified Answer
The correct answer is:
$M g(\sqrt{2}-1)$
Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculated by using work-energy theorem. Let us assume that body is taken slowly so that its speed doesn't change, then
$\begin{aligned}
\Delta K & =0 \\
& =W_F+W_{M g}+W_{\text {tension }}
\end{aligned}$
[symbols have their usual meanings]
$\begin{aligned}
W_F & =F \times l \sin 45^{\circ}, \\
W_{M g} & =M_g\left(l-l \cos 45^{\circ}\right), W_{\text {tension }}=0 \\
\therefore \quad F & =M g(\sqrt{2}-1)
\end{aligned}$
$\begin{aligned}
\Delta K & =0 \\
& =W_F+W_{M g}+W_{\text {tension }}
\end{aligned}$
[symbols have their usual meanings]
$\begin{aligned}
W_F & =F \times l \sin 45^{\circ}, \\
W_{M g} & =M_g\left(l-l \cos 45^{\circ}\right), W_{\text {tension }}=0 \\
\therefore \quad F & =M g(\sqrt{2}-1)
\end{aligned}$
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