Physics GRE GR 1777 Problem 046

046. Classical Mechanics (Hooke’s law)

$\vec{F} = - k \Delta x \ \ \text{where } k \text{ is a force constant and } x \text{ is a displacement (change in length of a spring).}$

$U = \frac{1}{2} k x^2 = \frac{1}{2} (kx) x = \frac{1}{2} |\vec{F}| x$

The same external force is applied to both springs,

$|\vec{F}| = k\Delta x = k_1 \Delta x_1 = k_2 \Delta x_2$

$\text{Since } k_1 > k_2, \text{ then } \Delta x_1 < \Delta x_2$

$U_1 = \frac{1}{2} k_1 (\Delta x_1)^2 = \frac{1}{2} (k_1 \Delta x_1) \Delta x_1 = \frac{1}{2} |\vec{F}| \Delta x_1$

$U_2 = \frac{1}{2} k_2 (\Delta x_2)^2 = \frac{1}{2} (k_2 \Delta x_2) \Delta x_2 = \frac{1}{2} |\vec{F}| \Delta x_2$

$\text{Since } \Delta x_1 < \Delta x_2, \text{ then } U_1 < U_2$

(A) $\text{Extension: } \Delta x_1 < \Delta x_2 \text{, Stored Potential Energy } U_1 < U_2$