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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

A block of mass MM is attached to the lower end of a vertical spring. The spring is hung from the ceiling and has a force constant value of kk. The mass is released from rest with the spring initially unstretched. The maximum extension produced along the length of the spring will be:

A

Mg/k

B

2Mg/k

C

4Mg/k

D

Mg/2k

Step-by-Step Solution

  1. Identify the Principle: Since the only forces doing work are gravity and the spring force (both conservative), the total mechanical energy of the system is conserved [Class 11 Physics, Ch 6, Sec 5.8].
  2. Initial State: The mass is released from rest (vi=0v_i = 0) from the unstretched position (x=0x = 0). Let the reference level for gravitational potential energy be at this initial position. Thus, Ki=0K_i = 0, Ug,i=0U_{g,i} = 0, Us,i=0U_{s,i} = 0. Total Initial Energy Ei=0E_i = 0.
  3. Final State (Maximum Extension): The mass descends a distance xmx_m (maximum extension). At this point, it momentarily comes to rest (vf=0v_f = 0), so kinetic energy Kf=0K_f = 0. The spring is stretched by xmx_m, so Elastic Potential Energy Us,f=12kxm2U_{s,f} = \frac{1}{2}kx_m^2 [Class 11 Physics, Ch 6, Sec 5.9, Eq 5.15]. The mass is at a height xm-x_m relative to the reference, so Gravitational Potential Energy Ug,f=MgxmU_{g,f} = -Mgx_m.
  4. Apply Conservation of Energy: Ei=EfE_i = E_f 0=12kxm2Mgxm0 = \frac{1}{2}kx_m^2 - Mgx_m Mgxm=12kxm2Mgx_m = \frac{1}{2}kx_m^2 Since xm0x_m \neq 0, we can divide by xmx_m: Mg=12kxmMg = \frac{1}{2}kx_m xm=2Mgkx_m = \frac{2Mg}{k} (Note: This is different from the equilibrium position where xeq=Mg/kx_{eq} = Mg/k. The mass oscillates around this equilibrium point).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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