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## Q. 2.7

Calculate the total potential energy of the system of Figure 2.15 as the mass is displaced a distance x downward form the system’s equilibrium position. Use a horizontal plane through the mass when the system is in equilibrium as the datum.

## Verified Solution

When the system is in equilibrium, the spring has a static deflection, $\Delta =\frac{mg}{k}$. Thus, as the mass moves down a distance x from the equilibrium position, the potential energy in the spring is

$V=\frac{1}{2}k\left(x+\Delta \right)^{2}$          (a)

Adding to this, the potential energy due to gravity Vgmgx yields

$V=\frac{1}{2}k\left(x+\Delta \right)^{2} -mgx$

$=\frac{1}{2}k\left(x+ =\frac{mg}{k} \right)^{2}-mgx$

$=\frac{1}{2}\left(kx^{2}-2mgx+\frac{m^{2}g^{2}}{k}\right) -mgx$

$=\frac{1}{2}kx^{2}+ V_{0}$           (b)

where $V_{0}=\frac{m^{2}g^{2}}{2k}$ is the potential energy in the spring when the system is in equilibrium. Thus, the total potential energy is expressed as the potential energy of the spring with respect to the equilibrium position plus the potential energy of the system when it is in equilibrium.