If a spring has a constant of 7 (kg)/s^2, how much work will it take to extend the spring by 22 cm ?

Apr 30, 2018

Given,

k = (7"N")/"s"

$\Delta x = 0.22 \text{m}$

Recall how external force is related to these variables,

${F}_{\text{ext}} = k x$

, and let's use the average force in determining the work done over displacement $\Delta x$, such that,

${W}_{\text{ext}} = \overline{F} \Delta x = \left(\frac{1}{2} k x\right) \cdot x = \frac{1}{2} k {x}^{2}$

This is familiar to elastic potential energy, which we are essentially working against in the positive direction. Hence, the work required to do this is,

${W}_{\text{ext" = 1/2kx^2 approx 0.17"J}}$