Rotational Motion

JEE Main level, single correct, medium

150 of 280

Question 150 of 280


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 :

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Answer: (D) 2 m2 \mathrm{~m}

wg+wFr+ws=ΔKE5×10×5−0.5×5×10×x−12Kx2=0−0250=25x+50x22x2+x−10=0x=2\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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