Question

A 4 kg mass is suspended as shown in the figure. All pulleys are frictionless and spring constant \( K \) is \( 8 \times 10^3 \) Nm\(^{-1}\). The extension in spring is ( \( g = 10 \) ms\(^{-2}\) )

• A. \( 2 \) mm
• B. \( 2 \) cm
• C. \( 4 \) cm
• D. \( 4 \) mm

Hint:

When a mass is attached to a spring in a pulley system, the force gets divided based on the configuration of the pulleys, which effectively reduces the force acting on the spring.

Answer 1

The Correct Option is B

Step 1: Analyze the forces on the system
The 4 kg mass exerts a downward force due to gravity: \[ F = mg = 4 \times 10 = 40 \text{ N} \] Since the pulleys are frictionless, the force is evenly distributed between the two segments of the spring, effectively creating an equivalent force of: \[ F_{\text{effective}} = \frac{40}{2} = 20 \text{ N} \] Step 2: Use Hooke's Law
Hooke's law states that: \[ F = K x \] Substituting the given values: \[ 20 = (8 \times 10^3) x \] Step 3: Solve for extension
\[ x = \frac{20}{8 \times 10^3} \] \[ x = \frac{20}{8000} = 0.0025 \text{ m} = 2 \text{ cm} \] Step 4: Conclusion
Thus, the extension in the spring is: \[ \boxed{2 \text{ cm}} \]