Question

A 3 kg block is connected as shown in the figure. Spring constants of two springs \( K_1 \) and \( K_2 \) are 50 Nm\(^{-1}\) and 150 Nm\(^{-1}\) respectively. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is ( \( g = 10 \) ms\(^{-2}\) )

• A. \( 10 \) ms\(^{-2}\)
• B. \( 12 \) ms\(^{-2}\)
• C. \( 8 \) ms\(^{-2}\)
• D. \( 8.8 \) ms\(^{-2}\)

Hint:

The acceleration of a block in a vertical spring system is determined using Hooke’s law and equilibrium conditions. The net force on the block at the lowest point gives the maximum acceleration.

Answer 1

The Correct Option is A

Step 1: Understanding the motion of the block
When the block reaches its lowest position, both the springs are stretched, and the restoring force exerted by the springs is given by: \[ F = (K_1 + K_2) x \] The force due to gravity acting on the block is: \[ F_{\text{gravity}} = mg = 3 \times 10 = 30 \text{ N} \] Step 2: Equilibrium condition
At the lowest point, the net restoring force is: \[ F_{\text{restoring}} = (50 + 150) x = 200x \] Since the block was initially at rest, the displacement \( x \) is determined using force balance: \[ 200x = 30 \] \[ x = \frac{30}{200} = 0.15 \text{ m} \] Step 3: Maximum acceleration
Acceleration at the lowest point is given by: \[ a = \frac{(K_1 + K_2) x}{m} \] \[ a = \frac{200 \times 0.15}{3} \] \[ a = \frac{30}{3} = 10 \text{ ms}^{-2} \] Step 4: Conclusion
Thus, the acceleration of the block in its lowest position is: \[ \boxed{10 \text{ ms}^{-2}} \]