Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A block of mass $m$ is moving with initial velocity $u$ towards a stationary spring of stiffness constant $k$ attached to the wall as shown in the figure. Maximum compression of the spring is: (The friction between the block and the surface is negligible).

$u\sqrt{\frac{m}{k}}$

$4u\sqrt{\frac{m}{k}}$

$2u\sqrt{\frac{m}{k}}$

$\frac{1}{2}u\sqrt{\frac{k}{m}}$

Step-by-Step Solution

Identify the Principle: Since friction is negligible, the total mechanical energy of the system is conserved. The initial kinetic energy of the block is converted into the elastic potential energy of the spring at maximum compression.
Formulate Equations:

Initial Kinetic Energy ( $K_i$ ) = $\frac{1}{2}mu^2$
Final Potential Energy ( $U_f$ ) at maximum compression $x_m$ = $\frac{1}{2}kx_m^2$ [Class 11 Physics, Ch 5, Sec 5.9, Eq 5.15].
At maximum compression, the block momentarily comes to rest, so final kinetic energy is zero.

Apply Conservation of Energy: $K_i = U_f$ $\frac{1}{2}mu^2 = \frac{1}{2}kx_m^2$
Solve for Maximum Compression ( $x_m$ ): $mu^2 = kx_m^2$ $x_m^2 = \frac{m}{k}u^2$ $x_m = u\sqrt{\frac{m}{k}}$ [Class 11 Physics, Ch 5, Example 5.8].

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started