A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is $200 \mathrm{~N} / \mathrm{m}$. The block is pushed such that the length of the spring becomes 1 m and then released. At distance $\mathrm{x} \mathrm{m}(\mathrm{x}<2)$ from the wall, the speed of the block will be:

(B) $10\left[1-(2-x)^{2}\right]^{1 / 2} \mathrm{~m} / \mathrm{s}$

(A) $10[1-(2-\mathrm{x})]^{3 / 2} \mathrm{~m} / \mathrm{s}$

(C) $10\left[1-(2-x)^{2}\right] \mathrm{~m} / \mathrm{s}$

(D) $10\left[1-(2-x)^{2}\right]^{2} \mathrm{~m} / \mathrm{s}$

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Work, Power & Energy

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is

200 N / m

. The block is pushed such that the length of the spring becomes 1 m and then released. At distance

xm (x < 2)

from the wall, the speed of the block will be:

(A) $10 [1 - (2 - x)]^{3/2} m / s$
(B) $10 [1 - (2 - x)^{2}]^{1/2} m / s$ ✓ Correct
(C) $10 [1 - (2 - x)^{2}] m / s$
(D) $10 [1 - (2 - x)^{2}]^{2} m / s$

Answer: (B)

Solution

On the frictionless table the initial elastic energy equals the kinetic energy plus the elastic energy at length

x

. The initial compression is

2 - 1 = 1 m

and the compression at distance

x

2 - x

, so

\frac{1}{2} K (1)^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} K (2 - x)^{2}

. With

K = 200

and

m = 2

this gives

v^{2} = 100 [1 - (2 - x)^{2}]

. The speed of the block is

10 [1 - (2 - x)^{2}]^{1/2} m/s

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Work, Power & Energy

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