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

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Elasticity & SHM

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 energy of the block at

x = 5

cm is 0.25 J. The spring constant of the spring is ____

Nm^{- 1}

Answer: 67

Released from rest at

x_{0} = 10

cm, the block's total mechanical energy is

\frac{1}{2} k x_{0}^{2}

, all stored in the spring. At

x = 5

cm the spring still holds

\frac{1}{2} k (0.05)^{2}

, and the remainder appears as the block's kinetic energy of

0.25

J. Energy conservation gives

\frac{1}{2} k x_{0}^{2} - \frac{1}{2} k (\frac{x _{0}}{2})^{2} = 0.25

, that is

\frac{1}{2} k (0.1)^{2} (1 - \frac{1}{4}) = 0.25

. Solving,

k = \frac{0.25}{\frac{1}{2} ( 0.01 ) ( 3/4 )} = \frac{200}{3}

. The spring constant is about

67 N m^{- 1}

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer