Example 2. What work is required to stretch a spring by $18 \mathrm{~cm}$, if a force of $24 \mathrm{H}$ stretches the spring by $3 \mathrm{~cm}$?
Solution. According to Hooke's Law, the elastic force stretching the spring is proportional to this extension, i.e., $F(x)=k x$. According to the condition: $F(0.03 \text{m})=24 \text{N}$. From the equation $24=0.03 k$ we find $k=800$, so $F(x)=800 x$ is the force stretching the spring. The work done by this force is

$$
A=\int_{0}^{0.18} 800 x d x=\left.400 x^{2}\right|_{0} ^{0.18}=25.92(\text{ J })
$$

Answer. 25.92 J.