Work — Question 10

Question 10 – Solution

1. Define displacement from natural length.

Let $x$ be extension measured positively downward. The spring lengths correspond to$$x_0=1-1=0,\qquad x_1=1.4-1=0.4\ \mathrm{m}.$$

2. Use the signed force exerted by the spring.

The spring pulls upward, opposite the positive direction. Therefore,$$F_s(x)=-kx=-300x\ \mathrm{N}.$$The question asks for work done by the spring itself.

3. Set up the signed work integral.

$$W_s=\int_0^{0.4}F_s(x)\,dx=\int_0^{0.4}-300x\,dx.$$

4. Integrate and apply the bounds.

$$W_s=\left[-150x^2\right]_0^{0.4}=-150(0.4)^2=-150(0.16)=-24.$$

5. State the result and interpret its sign.

$$\boxed{W_s=-24\ \mathrm{J}}.$$The negative sign indicates that the spring force opposes the motion. Its elastic potential energy increases by $\tfrac12k(0.4)^2=24$ J, consistent with $W_s=-\Delta U$.