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Physics Question 29 – JEE-MAIN 2025

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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

In a frictionless system with a spring, mechanical energy (kinetic + potential) is conserved.

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Ninja Strategy•Dimensional Analysis of Speed

Recognize that speed is derived from v2, so the final expression for v must involve a square root (power of 1/2) of the energy terms. Options with other powers are dimensionally incorrect for speed.

Step 1: Identify Initial and Final Energy States✦ Active

The system starts with the spring compressed to 1 m from its natural length of 2 m, meaning an initial compression of xi=(2−1)=1 m. The block is released from rest, so initial kinetic energy is zero. The initial potential energy stored in the spring is PEi=12kxi2. At a distance x from the wall, the spring's length is x. Since x<2 m, the spring is compressed by xf=(2−x) m. The final energy consists of kinetic energy KEf=12mv2 and potential energy PEf=12kxf2.

Step 2: Apply Conservation of Mechanical Energy○ Expand

According to the principle of conservation of mechanical energy, the total initial energy equals the total final energy (Ei=Ef). We have:

12mvi2+12kxi2=12mv2+12kxf2

Given m=2 kg, k=200 N/m, vi=0, xi=1 m, and xf=(2−x) m, substitute these values into the energy conservation equation:

12(2)(0)2+12(200)(1)2=12(2)v2+12(200)(2−x)2
Step 3: Solve for the speed v○ Expand

Simplify the equation to find v:

0+100=v2+100(2−x)2 v2=100−100(2−x)2 v2=100[1−(2−x)2] v=100[1−(2−x)2] v=10[1−(2−x)2]12m/s

This corresponds to option 3.

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