Question

A spring requires a force of 25 N to produce an extension of 2 cm in it. What is the work done in further extending it by 3 cm.
A. 1.3125 J
B. 1.876 J
C. 2.7121 J
D. 2.85 J

Answer 1

Question 32568

$$\Delta x _ {1} = 2 \mathrm{cm}; \Delta x _ {2} = 3 \mathrm{cm}; F _ {1} = 25 \mathrm{N}.$$

According to Hooke's law, $|F| = k\Delta x$, where $k$ is the characteristic of the spring.

For an extension $\Delta x_{1} = 2\mathrm{cm}$, $|F_{1}| = k\Delta x_{1}$, from which $k = \frac{F_1}{\Delta x_1}$.

For $\Delta x_{2} = 3\mathrm{cm}$, $|F_{2}| = k\Delta x_{2}$, and using expression for $k$ above, obtain

$$F _ {2} = \frac {F _ {1}}{\Delta x _ {1}} \cdot \Delta x _ {2} = 37.5 \mathrm{N}$$