Deformation of Solids: Question 5

Worked solution

Setting up the problem

The elastic potential energy stored in a spring at extension $x$ (within the region where Hooke’s law is obeyed) is: $E_p = \tfrac12 kx^2$

The additional energy stored between two extensions is found by calculating $E_p$ at each extension separately and subtracting. It is not the same as applying the formula to the difference in extension, because $E_p$ depends on $x^2$, not on $x$.

Converting the extensions to metres: $x_1 = 5.0\text{ cm} = 0.050\text{ m}, \qquad x_2 = 9.0\text{ cm} = 0.090\text{ m}$

Calculating the two energies

At $x_1 = 0.050\text{ m}$: $E_{p1} = \tfrac12(40)(0.050)^2 = \tfrac12(40)(0.0025) = 0.050\text{ J}$

At $x_2 = 0.090\text{ m}$: $E_{p2} = \tfrac12(40)(0.090)^2 = \tfrac12(40)(0.0081) = 0.162\text{ J}$

Finding the additional energy stored

$\Delta E_p = E_{p2} - E_{p1} = 0.162 - 0.050$ $\Delta E_p = 0.112\text{ J}$

Why the other options are wrong

• A ($0.032\text{ J}$): this comes from wrongly applying $\tfrac12 kx^2$ to the difference in extension, $x_2-x_1=0.040\text{ m}$, i.e. $\tfrac12(40)(0.040)^2$, which is not valid since elastic potential energy is not a linear function of extension.
• C ($0.162\text{ J}$): this is the total energy stored at $9.0\text{ cm}$ ($E_{p2}$), forgetting to subtract the energy already stored at $5.0\text{ cm}$.
• D ($0.224\text{ J}$): this comes from omitting the factor of $\tfrac12$, effectively calculating $kx_2^2 - kx_1^2 = 40(0.0081) - 40(0.0025) = 0.324-0.100=0.224\text{ J}$.

Final answer

• Additional elastic potential energy stored $= \boxed{0.112}\text{ J}$, option B.