Question

A spring of \( 5 \times 10^3 \) Nm\(^{-1} \) spring constant is stretched initially by 10 cm from the unstretched position. The work required to stretch it further by another 10 cm is

• A. 75 N-m
• B. 50 N-m
• C. 76 N-m
• D. 82 N-m

Hint:

For calculating work done in stretching a spring, always use the energy difference formula instead of just \( \frac{1}{2} k x^2 \) to avoid errors.

Answer 1

The Correct Option is A

The work done in stretching a spring is given by the elastic potential energy formula: \[ W = \frac{1}{2} k (x_f^2 - x_i^2) \] where \( k = 5 \times 10^3 \) Nm\(^{-1} \), \( x_i = 10 \) cm = 0.1 m, \( x_f = 20 \) cm = 0.2 m. Substituting the values: \[ W = \frac{1}{2} \times 5000 \times (0.2^2 - 0.1^2) \] \[ = \frac{1}{2} \times 5000 \times (0.04 - 0.01) \] \[ = \frac{1}{2} \times 5000 \times 0.03 \] \[ = \frac{5000 \times 0.03}{2} = \frac{150}{2} = 75 \text{ N-m} \] Thus, the required work is 75 N-m.