Source: High school physics (Chinese)
Problem Sets:
Problem
A spring with a natural length of $20$ cm and a spring constant of $k = 1000$ N/m is stretched.
- What is the increase in elastic potential energy when the spring is stretched from its natural length to $25$ cm?
- What is the increase in elastic potential energy when it is further stretched from $25$ cm to $30$ cm?
- Are these two increases in energy equal?
\Delta E_{p1} = 1.25 J \Delta E_{p2} = 3.75 J No, they are not equal.
The change in elastic potential energy is $\Delta E_p = \frac{1}{2}k(x_f^2 - x_i^2)$, where $x$ is the extension from the natural length.
[Q1] Stretching from $20$ cm to $25$ cm. The initial extension is $x_i = 0$ m and the final extension is $x_f = 25 \text{ cm} - 20 \text{ cm} = 5 \text{ cm} = 0.05$ m.
$$\Delta E_{p1} = \frac{1}{2}k(x_f^2 - x_i^2) = \frac{1}{2}(1000 \text{ N/m})((0.05 \text{ m})^2 - 0^2) = 1.25 \text{ J}$$[Q2] Stretching from $25$ cm to $30$ cm. The initial extension is $x_i = 5$ cm $= 0.05$ m and the final extension is $x_f = 30 \text{ cm} - 20 \text{ cm} = 10 \text{ cm} = 0.10$ m.
$$\Delta E_{p2} = \frac{1}{2}k(x_f^2 - x_i^2) = \frac{1}{2}(1000 \text{ N/m})((0.10 \text{ m})^2 - (0.05 \text{ m})^2) = 3.75 \text{ J}$$[Q3] Comparing the results, $\Delta E_{p1} eq \Delta E_{p2}$. The increases are not equal.