Question

A spring with a spring constant of 800 N/m is stretched from its natural length by 5 cm. The additional work required to stretch it from 5 cm to 10 cm is:

Answer 1

3 J

Answer 2

The work done to stretch a spring from an extension $x_1$ to $x_2$ is given by $W = \frac{1}{2} k (x_2^2 - x_1^2)$. Given $k = 800$ N/m, initial extension $x_1 = 5$ cm $= 0.05$ m, and final extension $x_2 = 10$ cm $= 0.10$ m. Substituting these values, $W = \frac{1}{2} \times 800 \times ((0.10)^2 - (0.05)^2)$. $W = 400 \times (0.01 - 0.0025) = 400 \times 0.0075 = 3$ J.