Q116 / 150 · MHT CET 2026 (13 April Shift 2) · Physics
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$65$
- Correct answerC$67$
- D$50$
Solution
The amplitude of the motion is $A = 10 \text{ cm} = 0.1 \text{ m}$.
The kinetic energy of a block in simple harmonic motion at a distance $x$ from the equilibrium position is given by $K = \dfrac{1}{2} k (A^2 - x^2)$.
Given $K = 0.25 \text{ J}$ at $x = 5 \text{ cm} = 0.05 \text{ m}$.
Substituting the values:
$0.25 = \dfrac{1}{2} k ((0.1)^2 - (0.05)^2)$
$0.5 = k (0.01 - 0.0025)$
$0.5 = k (0.0075)$
$k = \dfrac{0.5}{0.0075} = \dfrac{5000}{75} = \dfrac{200}{3} \approx 66.67 \text{ Nm}^{-1}$
The spring constant is nearly $67 \text{ Nm}^{-1}$.
Answer: $67$
The kinetic energy of a block in simple harmonic motion at a distance $x$ from the equilibrium position is given by $K = \dfrac{1}{2} k (A^2 - x^2)$.
Given $K = 0.25 \text{ J}$ at $x = 5 \text{ cm} = 0.05 \text{ m}$.
Substituting the values:
$0.25 = \dfrac{1}{2} k ((0.1)^2 - (0.05)^2)$
$0.5 = k (0.01 - 0.0025)$
$0.5 = k (0.0075)$
$k = \dfrac{0.5}{0.0075} = \dfrac{5000}{75} = \dfrac{200}{3} \approx 66.67 \text{ Nm}^{-1}$
The spring constant is nearly $67 \text{ Nm}^{-1}$.
Answer: $67$
Hence, the correct answer is option: 'C'.