382. A force of $60 \mathrm{H}$ stretches a spring by 2 cm. The original length of the spring is $14 \mathrm{~cm}$. How much work is required to stretch it to 20 cm?
Solution. We have $k=\frac{60}{0.02}=3000(\mathrm{H} /$ m) and, consequently, $F=$ $=3000 x$. Since the spring needs to be stretched by 0.06 (m), then

$$
A=\int_{0}^{0.06} 3000 x d x=\left.1500 x^{2}\right|_{0}^{0.06}=1500 \cdot 0.06^{2}=5.4 \quad(\text{J})
$$