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

Initially, the block has kinetic energy \( E \). When its speed is halved, its kinetic energy becomes \(\frac{1}{2}mv_f^2 = \frac{E}{2}\). The loss in kinetic energy is \( E - \frac{E}{2} = \frac{E}{2} \).
This energy is used to compress the spring. The potential energy stored in the spring is given by \(\frac{1}{2}kx^2\), where \( x = 0.25 \, \text{m} \) is the compression.
Equating the loss in kinetic energy to the potential energy stored in the spring, we have:
\[\frac{1}{2}kx^2 = \frac{E}{2}\]
Substituting \( x = 0.25 \, \text{m} \):
\[\frac{1}{2}k(0.25)^2 = \frac{E}{2}\]
\[\frac{1}{2}k \times 0.0625 = \frac{E}{2}\]
\[k \times 0.0625 = E\]
\[k = \frac{E}{0.0625} = 16E\]
Thus, the spring constant \( k \) is \( 16E \, \text{Nm}^{-1} \). The value of \( n \) is 16.
Checking against the range, \( n \) is within the required range of 24 to 24.