Work — Question 2

Question 2 – Solution

1. Use extension beyond natural length.

Let $x$ be extension, not total spring length. The calibration force produces$$x=0.8-0.5=0.3\ \mathrm{m}.$$

2. Find the spring constant.

Hooke’s law for the required stretching force is $F=kx$:$$12=k(0.3)\quad\Longrightarrow\quad k=\frac{12}{0.3}=40\ \mathrm{N/m}.$$

3. Translate the requested lengths into bounds.

The initial and final extensions are$$x_0=0.5-0.5=0,\qquad x_1=1.0-0.5=0.5\ \mathrm{m}.$$Thus $F(x)=40x$ and $W=\int_0^{0.5}40x\,dx$.

4. Integrate and evaluate.

$$W=\left[20x^2\right]_0^{0.5}=20(0.5)^2-0=20(0.25)=5.$$

5. State and check the work.

$$\boxed{W=5\ \mathrm{J}}.$$The force rises linearly from $0$ to $20$ N, so the triangular area is also $\tfrac12(0.5)(20)=5$ J.