Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

Mg/k

2Mg/k

4Mg/k

Mg/2k

Step-by-Step Solution

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].
Initial State: The mass is released from rest ( $v_i = 0$ ) from the unstretched position ( $x = 0$ ). Let the reference level for gravitational potential energy be at this initial position. Thus, $K_i = 0$ , $U_{g,i} = 0$ , $U_{s,i} = 0$ . Total Initial Energy $E_i = 0$ .
Final State (Maximum Extension): The mass descends a distance $x_m$ (maximum extension). At this point, it momentarily comes to rest ( $v_f = 0$ ), so kinetic energy $K_f = 0$ . The spring is stretched by $x_m$ , so Elastic Potential Energy $U_{s,f} = \frac{1}{2}kx_m^2$ [Class 11 Physics, Ch 6, Sec 5.9, Eq 5.15]. The mass is at a height $-x_m$ relative to the reference, so Gravitational Potential Energy $U_{g,f} = -Mgx_m$ .
Apply Conservation of Energy: $E_i = E_f$ $0 = \frac{1}{2}kx_m^2 - Mgx_m$ $Mgx_m = \frac{1}{2}kx_m^2$ Since $x_m \neq 0$ , we can divide by $x_m$ : $Mg = \frac{1}{2}kx_m$ $x_m = \frac{2Mg}{k}$ (Note: This is different from the equilibrium position where $x_{eq} = Mg/k$ . The mass oscillates around this equilibrium point).

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