Astationary mass m = 1.3 kg is hanging from a spring of spring constant k = 1200 n/m. you raise the mass a distance of 10 cm above its equilibrium position. how much has the potential energy of the mass-spring system changed?

The change in the potential energy of the mass-spring system is 7.27 J.

The given parameters;

mass, m = 1.3 kgspring constant, k = 1200 N/mheight the mass is raised, h = 10 cm = 0.1 m

The change in the potential energy of the mass-spring system is calculated as follows;

total change in potential energy =  change in potential energy due to height of the mass  +  change in potential energy due to compression of the spring.

Thus, the change in the potential energy of the mass-spring system is 7.27 J.

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The weight of the mass is

This force stretches the spring of a certain amount , which can be calculated by using Hook's law:

with respect to the equilibrium position of the spring.
So, the initial elastic potential energy of the spring is


Later, the spring is compressed by  from its equilibrium position (the negative sign means that now it is a compression, while before it was an elongation). This means that the new elastic potential energy of the spring is


So, the change in elastic potential energy of the mass-spring system is


However, if we consider also the change in gravitational potential energy of the mass, the answer is different. In fact, the height of the mass changed by 

So its gravitational potential energy increased by


So, the total change of potential energy (elastic+gravitational) of the mass-spring system is

The correct answer is c. hope this