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Check Your Understanding When the length of the spring in [link] changes from an initial value of 22.0 cm to a final value, the elastic potential energy it contributes changes by −0.0800 J . Find the final length.

22.8 cm. Using 0.02 m for the initial displacement of the spring (see above), we calculate the final displacement of the spring to be 0.028 m; therefore the length of the spring is the unstretched length plus the displacement, or 22.8 cm.

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Gravitational and elastic potential energy

A simple system embodying both gravitational and elastic types of potential energy is a one-dimensional, vertical mass-spring system . This consists of a massive particle (or block), hung from one end of a perfectly elastic, massless spring, the other end of which is fixed, as illustrated in [link] .

A vertical mass spring system is illustrated. The upper end of the spring is attached to the ceiling. A block of mass m is attached to the lower end.  The spring is drawn at two positions. On the left, the mass is in the equilibrium position. To  the right of this, the spring is drawn with the mass pulled down a distance y sub pull. This position of the mass is labeled as h equal to zero. A graph of y as a function of X is shown to the rightly the illustrations, with y equals zero aligned with the equilibrium position in the illustrations. The plot is sinusoidal, with the minimum y at x=0 and even with the lower mass position in the illustrations.
A vertical mass-spring system, with the y -axis pointing upwards. The mass is initially at an equilibrium position and pulled downward to y pull . An oscillation begins, centered at the equilibrium position.

First, let’s consider the potential energy of the system. Assuming the spring is massless, the system of the block and Earth gains and loses potential energy. We need to define the constant in the potential energy function of [link] . Often, the ground is a suitable choice for when the gravitational potential energy is zero; however, in this case, the lowest point or when h = 0 is a convenient location for zero gravitational potential energy. Note that this choice is arbitrary, and the problem can be solved correctly even if another choice is picked.

We must also define the elastic potential energy of the system and the corresponding constant, as detailed in [link] . The equilibrium location is the most suitable mathematically to choose for where the potential energy of the spring is zero.

Therefore, based on this convention, each potential energy and kinetic energy can be written out for three critical points of the system: (1) the lowest pulled point, (2) the equilibrium position of the spring, and (3) the highest point achieved. We note that the total energy of the system is conserved, so any total energy in this chart could be matched up to solve for an unknown quantity. The results are shown in [link] .

Components of energy in a vertical mass-spring system
Gravitational P.E. Elastic P.E. Kinetic E.
(3) Highest Point 2 m g y pull 1 2 k y 2 pull 0
(2) Equilibrium m g y pull 0 1 2 m v 2
(1) Lowest Point 0 1 2 k y 2 pull 0
A photograph of a bungee jumper.
A bungee jumper transforms gravitational potential energy at the start of the jump into elastic potential energy at the bottom of the jump.

Potential energy of a vertical mass-spring system

A block weighing 12 N is hung from a spring with a spring constant of 6.0 N / m , as shown in [link] . The block is pulled down an additional 5.0 cm from its equilibrium position and released. (a) What is the difference in just the spring potential energy, from an initial equilibrium position to its pulled-down position? (b) What is the difference in just the gravitational potential energy from its initial equilibrium position to its pulled-down position? (c) What is the kinetic energy of the block as it passes through the equilibrium position from its pulled-down position?

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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