Question

A block of mass ‘m’ (as shown in figure) moving with kinetic energy E compresses a spring through a distance 25 cm when its speed is halved. The value of spring constant of used spring will be nE Nm^–1 for n = _______.

Answer 1

Correct Answer: 24

To determine the spring constant \(k\), we need to analyze the energy transformations involved. Initially, the kinetic energy of the block is \(E\). When the speed of the block is halved, its kinetic energy becomes \(\frac{1}{2} \times \frac{1}{2}mv^2 = \frac{E}{2}\). Thus, the kinetic energy \(\Delta E\) lost is \(E - \frac{E}{2} = \frac{E}{2}\).

The lost kinetic energy is converted into potential energy stored in the spring:

\[\frac{1}{2}kx^2 = \frac{E}{2}\]

Given \(x = 0.25 \, \text{m}\), substituting we have:

\[\frac{1}{2}k(0.25)^2 = \frac{E}{2}\]

\[k \times 0.0625 = E\]

\[k = \frac{E}{0.0625}\]

Calculating the value:

\[k = 16E \, \text{Nm}^{-1}\]

Thus, \(n = 16\) which falls within the specified range of 24,24, confirming the calculated spring constant is accurate.

Answer 2

When the velocity is halved the kinetic energy is reduced by a factor of \(\frac{1}{4}\).
Δ_KE= W all So \(\frac{E}{4-E}\)=−\(\frac{1}{2}\)K×(0.25)²×\(\frac{3E}{4}\)=\(\frac{1}{2}\)K×\(\frac{1}{16}\)K=24 E