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: Understanding the Concept:
The block is released from rest at \(x = 10\) cm, so the amplitude is \(A = 10\) cm. At displacement \(x\), the total energy is \(\frac12kA^2\), and the potential energy is \(\frac12kx^2\). So the kinetic energy is \(\frac12k(A^2-x^2)\).

Step 2: Calculate:
\(A = 0.10\) m and \(x = 0.05\) m. \(A^2 - x^2 = 0.01 - 0.0025 = 0.0075\) m\(^2\).
\[ 0.25 = \frac12k(0.0075) \Rightarrow k = \frac{0.5}{0.0075} = 66.7\ \text{N/m} \approx 67\ \text{N/m} \]

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