Physics GRE GR 1777, Problem 046 Solution

Physics GRE GR 1777 Problem 046

046. Classical Mechanics (Hooke’s law)

You should know about

\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

Solution

  • Extension

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

 

  • Stored Potential Energy

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

Answer

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

References

https://en.wikipedia.org/wiki/Hooke%27s_law

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