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: Energy Conservation 

When the block is released from rest, all of its gravitational potential energy is converted into spring potential energy at its lowest point. Let both springs stretch by \( x \) meters (since they are symmetrically placed and the block moves straight down).

Step 2: Spring and Gravitational Energy

Total spring potential energy: \[ PE_{\text{spring}} = \frac{1}{2}K_1 x^2 + \frac{1}{2}K_2 x^2 = \frac{1}{2}(K_1 + K_2)x^2 = \frac{1}{2}(50 + 150)x^2 = 100x^2 \] Gravitational potential energy lost: \[ PE_{\text{gravity}} = m g x = 3 \cdot 10 \cdot x = 30x \]

Step 3: Apply Energy Conservation

\[ 30x = 100x^2 \quad \Rightarrow \quad 3x = 10x^2 \quad \Rightarrow \quad x = 0.3 \, \text{m} \]

Step 4: Net Force at Lowest Point

At the lowest point, net force acting on the block: \[ F = mg - (K_1 + K_2)x = 30 - (50 + 150)\cdot 0.3 = 30 - 200 \cdot 0.3 = 30 - 60 = -30 \, \text{N} \] The negative sign indicates the net force is upward (restoring direction).

Step 5: Acceleration

\[ a = \frac{F}{m} = \frac{-30}{3} = -10 \, \text{m/s}^2 \] Taking magnitude (as the question asks for acceleration, not direction): \[ \boxed{a = 10 \, \text{m/s}^2} \]

Conclusion:

The acceleration of the block at its lowest point is \( \boxed{10 \, \text{m/s}^2} \).

Answer 2

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: \[ \boxed{10 \text{ ms}^{-2}} \]