JEE Main 2025 Apr 8 Shift 2, Physics Q20: Work Energy Theorem And Its Applications

JEE Main2025Apr 8, Shift 2Physics

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 $x$ m ( $x < 2$ ) 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
C
$10 [1 - (2 - x)^{2}]$ m/s
D
$10 [1 - (2 - x)^{2}]^{2}$ m/s

Solution

Initial compression when spring length is 1 m: $x_{i} = 2 - 1 = 1$ m.

When the block is at distance $x$ from the wall (with $x < 2$), the spring is still compressed by $x_f = 2 - x$ m.

Energy conservation: initial KE $+$ initial spring PE $=$ final KE $+$ final spring PE, i.e. $0 + \frac{1}{2} k x_{i}^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} k x_{f}^{2}$ .

Substituting $m = 2$ , $k = 200$ , $x_{i} = 1$ , $x_{f} = 2 - x$ : $\frac{1}{2} \cdot 200 \cdot 1 = \frac{1}{2} \cdot 2 \cdot v^{2} + \frac{1}{2} \cdot 200 \cdot (2 - x)^{2}$ .

Simplifying: $100 = v^{2} + 100 (2 - x)^{2}$ , so $v^{2} = 100 [1 - (2 - x)^{2}]$ .

Taking square root: $v = 10 1 - (2 - x)^{2}$ m/s, matching option (2).

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JEE Main 2025 Apr 8 Shift 2, Physics Q20: Work Energy Theorem And Its Applications

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