Question

A block is fastened to a horizontal spring. The block is pulled to a distance \(x = 10\) cm from its equilibrium position (at \(x = 0\)) on a frictionless surface from rest. The kinetic energy of the block at \(x = 5\) cm is \(0.25\) J. The spring constant of the spring is nearly (in \(\text{Nm}^{-1}\))

• A. \(63\)
• B. \(65\)
• C. \(67\)
• D. \(50\)

Hint:

In SHM, \(KE = \frac12k(A^2-x^2)\).

Answer 1

The Correct Option is C

Step 1: Plan:
Use energy conservation between the release point and the $x = 5$ cm point.

Step 2: Steps:
Energy at release: $\frac12k(0.1)^2 = 0.005k$. At $x = 5$ cm: $0.25 + \frac12k(0.05)^2 = 0.25 + 0.00125k$.
Equate: $0.005k - 0.00125k = 0.25$, so $0.00375k = 0.25$ and $k = 66.7$ N/m, about $67$ N/m.

Final Answer:
The spring constant is about $67$ N/m, option (C). \[ \boxed{67\ \text{N/m}} \]