JEE Mains 2025 Spring-Mass System Speed Problem Physics | | 8th April Shift 2
In this blog post, we tackle a conceptual numerical from the JEE Mains 2025 Physics paper held on 8th April (Shift 2). The question involves a spring-mass system on a frictionless surface, testing your understanding of energy conservation and spring potential energy. Let’s dive into the solution!
🎥 Watch the Full Video Solution
This video explains the complete step-by-step solution. Watch it first before reading the breakdown below.
Question
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:
- 10 × [1 − (2 − x)2]2 m/s
- 10 × [1 − (2 − x)2]1/2 m/s
- 10 × [1 − (2 − x)]3/2 m/s
- 10 × [1 − (2 − x)]2 m/s
More JEE Mains 2025 Questions
Detailed Step-by-Step Solution
Step 1: Identify the Process
The system involves a spring and a mass moving on a frictionless table, so we’ll use the principle of conservation of mechanical energy.
Step 2: Initial Energy of the System
Initially, the spring is compressed to a length of 1 m. The spring’s natural length is 2 m, so the compression is 1 m.
Initial potential energy of the spring:
Ui = (1/2) k x² = (1/2) × 200 × (1)² = 100 J
Step 3: Final Energy of the System
At a general position x (x < 2), the spring is stretched by (2 − x) meters. So, the potential energy becomes:
Uf = (1/2) × 200 × (2 − x)² = 100(2 − x)² J
Let the speed of the block at that point be v. Then, kinetic energy = (1/2)mv² = (1/2) × 2 × v² = v²
Step 4: Apply Conservation of Mechanical Energy
Total energy remains constant, so:
Initial PE = Final PE + KE
100 = 100(2 − x)² + v²
Simplifying:
v² = 100 [1 − (2 − x)²]
v = 10 × √[1 − (2 − x)²]
Step 5: Match with Given Options
The expression matches exactly with option (2):
10 × [1 − (2 − x)²]1/2 m/s
Conclusion
Using conservation of energy, we determined that the speed of the block at any position x < 2 m is given by v = 10 × √[1 − (2 − x)²]. Option (2) is the correct answer.
