Worked example 5.3: Stretching a spring

Question: The force required to slowly stretch a spring varies from 0N to 105N as the spring is extended by 13cm from its unstressed length. What is the force constant of the spring? What work is done in stretching the spring? Assume that the spring obeys Hooke's law.

Answer: The force $f$ that the spring exerts on whatever is stretching it is $f=-k x$, where $k$ is the force constant, and $x$ is the extension of the spring. The minus sign indicates that the force acts in the opposite direction to the extension. Since the spring is stretched slowly, the force $f'$ which must be exerted on it is (almost) equal and opposite to $f$. Hence, $f' = -f = k x$. We are told that $f'= 105$N when $x=0.13$m. It follows that

$$k = \frac{105}{0.13} = 807.7 {\rm N/m}.$$

The work $W'$ done by the external force in extending the spring from 0 to $x$ is

$$W' = \int_0^x f' dx = k\int_0^x x dx = \frac{1}{2} k x^2.$$

Hence,

$$W' = 0.5\times 807.7\times 0.13^2 = 6.83 {\rm J}.$$