Deformation of Solids: Question 10

Worked solution

Part (a): Extensions and the spring constant

The extension at each load is the total length minus the unstretched length of $12.0\text{ cm}$:

Checking the ratio $x/F$ for the first four rows: $2.5/1.0 = 5.0/2.0 = 7.5/3.0 = 2.5\text{ cm N}^{-1}$ in each case, a constant ratio, confirming direct proportionality (Hooke’s law) up to $F=3.0\text{ N}$.

Using, for example, $F=3.0\text{ N}$ with $x = 7.5\text{ cm} = 0.075\text{ m}$: $k = \frac{F}{x} = \frac{3.0}{0.075}$ $k = 40\text{ N m}^{-1}$

Part (b): Where Hooke’s law breaks down

Extrapolating the proportional relationship found in (a), the extension expected at $F=4.0\text{ N}$ would be: $x_{\text{expected}} = \frac{F}{k} = \frac{4.0}{40} = 0.10\text{ m} = 10.0\text{ cm}$

The actual measured extension at $F=4.0\text{ N}$ is $13.0\text{ cm}$, which is greater than the $10.0\text{ cm}$ predicted by direct proportion. This shows that $F$ and $x$ are no longer in direct proportion once the load reaches $4.0\text{ N}$, so the spring has exceeded the limit of proportionality between $3.0\text{ N}$ and $4.0\text{ N}$.

Part (c): Elastic potential energy at $3.0\text{ N}$

Since $F=3.0\text{ N}$ is within the region where the spring obeys Hooke’s law, the elastic potential energy stored is: $E_p = \tfrac12 F x$

Using $F=3.0\text{ N}$ and $x = 7.5\text{ cm} = 0.075\text{ m}$: $E_p = \tfrac12 \times 3.0 \times 0.075$ $E_p = 0.1125\text{ J} \approx 0.113\text{ J}$

As a check, using $E_p = \tfrac12 k x^2$: $E_p = \tfrac12 \times 40 \times (0.075)^2 = \tfrac12 \times 40 \times 0.005625 = 0.1125\text{ J}$ Both methods agree.

Final answers

• (a) Extensions: $0, 2.5, 5.0, 7.5, 13.0\text{ cm}$; spring constant $k = \boxed{40}\text{ N m}^{-1}$
• (b) Hooke’s law first breaks down at $\boxed{4.0}\text{ N}$, since the extension exceeds the value predicted by direct proportion.
• (c) Elastic potential energy at $3.0\text{ N}$ $= \boxed{0.113}\text{ J}$