JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 2)
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A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is :
$$\sqrt{6} \mathrm{~m}$$
$$\sqrt{5} \mathrm{~m}$$
$$1 \mathrm{~m}$$
$$2 \mathrm{~m}$$
Explanation
$\begin{aligned} & \mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{Fr}}+\mathrm{W}_{\mathrm{s}}=\Delta \mathrm{KE} \\\\ & 5 \times 10 \times 5-0.5 \times 5 \times 10 \times \mathrm{x}-\frac{1}{2} \mathrm{Kx}^2=0-0 \\\\ & 250=25 \mathrm{x}+50 \mathrm{x}^2 \\\\ & 2 \mathrm{x}^2+\mathrm{x}-10=0 \\\\ & \mathrm{x}=2\end{aligned}$
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