Physics · Work, Power & Energy

JEE Main 2024 — 8 April, Shift 2 — Question 29

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 : (Neglect any energy losses during collision with the ground)

Option A:
$\sqrt{6} \mathrm{~m}$
Option B:
2 m
Correct
Option C:
1 m
Option D:
$\sqrt{5} \mathrm{~m}$

Answer: B

Step-by-step solution

$\mathrm{w}_{\mathrm{g}}+\mathrm{w}_{\mathrm{Fr}}+\mathrm{w}_{\mathrm{s}}=\Delta \mathrm{KE}$

$5 \times 10 \times 5-0.5 \times 5 \times 10 \times x-\frac{1}{2} K x^{2}=0-0$ $250=25 x+50 x^{2}$

$2 \mathrm{x}^{2}+\mathrm{x}-10=0$ $\mathrm{x}=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Work, Power & Energy
Topic: Applications of Conservation of Mechanical Energy