Question

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 49.1 N, the spring is stretched by 15.2 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 13.7 cm from that position.

Answer 1

When the spring is stretched by 15.2 cm = 0.152 m, the spring exerts a restorative force with magnitude (due to Hooke's law)

$F = kx$

where $k$ is the spring constant. Solve for $k$.

$49.1\,\mathrm N = k (0.152\,\mathrm m) \implies k \approx 323 \dfrac{\rm N}{\rm m}$

The amount of work required to stretch or compress a spring by $x\,\mathrm m$ from equilibrium length is

$W = \dfrac12 kx^2$

Then the work needed to stretch the spring by 15.2 cm is

$W_1 = \dfrac12 \left(343\dfrac{\rm N}{\rm m}\right) (0.152\,\mathrm m)^2 \approx 3.73\,\mathrm J$

and by 15.2 + 13.7 = 28.9 cm is

$W_2 = \dfrac12 \left(343\dfrac{\rm N}{\rm m}\right) (0.289\,\mathrm m)^2 \approx 13.5\,\mathrm J$

so the work needed to stretch from 15.2 cm to 28.9 cm from equilibrium is

$\Delta W = W_2 - W_1 \approx \boxed{9.76\,\mathrm J}$