Topic: Engineering Mechanics
The difference in the Potential energies between two arbitrary positions downward displacement $X_{1}$ and $X_{2}$ ($X_{2}>X_{1}$).
- A. $\frac{1}{2}k[x_{1}-x_{2}]^{2}$
- B. $\frac{1}{2}k[(x_{2})^{2}-(x_{1})^{2}]$
- C. $\frac{1}{2}k[(x_{1})-(x_{2})]$
- D. $\frac{1}{2}k[x_{1}-x_{2}]$
Answer: B. $\frac{1}{2}k[(x_{2})^{2}-(x_{1})^{2}]$
Explanation: The potential energy of a spring at a downward displacement $x_{1}$ is $P_{1} = \frac{1}{2}kx_{1}^{2}$. The potential energy at displacement $x_{2}$ is $P_{2} = \frac{1}{2}kx_{2}^{2}$. Since $x_{2} > x_{1}$, $P_{2} > P_{1}$. The difference in potential energies is $P_{2} - P_{1} = \frac{1}{2}kx_{2}^{2} - \frac{1}{2}kx_{1}^{2} = \frac{1}{2}k[(x_{2})^{2} - (x_{1})^{2}]$.