Imagine a rock is shot from a slingshot and strikes the bullseye on a hanging target. What gave the rock motion? The elastic potential energy from the rubber bands is converted into kinetic energy as the rock leaves the slingshot and flies through the air. In this article, we will define elastic potential energy and discuss the formula for elastic potential energy of a spring. We will then go over an example to practice finding the elastic potential energy of a system.
Definition of Elastic Potential Energy
In the article, "Potential Energy and Energy Conservation", we discuss how potential energy is related to the internal configuration of an object. The elasticity of an object is part of its internal configuration that affects the energy of a system. Some objects, like rubber bands or springs, have a high elasticity, which means that the object can be stretched or compressed a significant amount and then go back to its original form after deformation. When an object is stretched or compressed, it stores elastic potential energy which can be used later.
Elastic potential energy: energy that is stored in an elastic object, like a rubber band or a spring, and can be used later
Units of Elastic Potential Energy
Elastic potential energy has the same units as all other forms of energy. The SI unit of energy is the joule, \(\mathrm{J}\), and is equivalent to a newton-meter so that \(\mathrm{J} = \mathrm{N}\,\mathrm{m}\).
Formula for Elastic Potential Energy
For potential energy in general, the change in the potential energy of a system is proportional to the work done by a conservative force. So for an elastic object, we find the formula for the elastic potential energy by considering the work the elastic object can do once compressed or stretched. In this article, we will focus on the elastic potential energy of a spring.
The spring force pulls a spring back to its equilibrium position, Vaia Originals
Hooke's law tells us that the force required to keep a spring stretched a distance, \(x\), from its natural position is given by \(F=kx\), where \(k\) is the spring constant that tells us how stiff the spring is. The image above shows a block on a spring being stretched with a force, \(F_p\), and then compressed with the same force. The spring pulls back with force \(F_s\) of the same magnitude in a direction opposite to that of the applied force. We do positive work on the spring by stretching or compressing it while the spring does negative work on us.
The work done on the spring to bring it into the stretched position is the force multiplied by the distance it is stretched. The magnitude of the spring force changes with respect to the distance, so let's consider the average force that it takes to stretch the spring over that distance. The average force required to stretch a spring from its equilibrium position, \(x=0\,\mathrm{m}\), to a distance, \(x\), is given by
$$ \begin{aligned} F_{avg} &= \frac{1}{2}\left(0\,\mathrm{m} + kx\right) \\ &= \frac{1}{2}kx \end{aligned}$$.
Then, the work done to stretch the spring is
$$ \begin{aligned} W &= F_{avg}x \\ &= \left(\frac{1}{2}kx\right)x \\ &= \frac{1}{2}kx^2 \end{aligned}$$.
Elastic Potential Energy Equation for a Spring
We have found the work done to stretch the spring from equilibrium to a certain distance, and the work is proportional to the change in elastic potential energy. The initial elastic potential energy is zero at the equilibrium position, so the equation for the elastic potential energy of a stretched spring is:
$$ U_{el} = \frac{1}{2}kx^2 $$
Since the distance is squared, for a negative distance, like when compressing a spring, the elastic potential energy is still positive.
Notice that the zero-point for elastic potential energy is the position at which the spring is at equilibrium. With gravitational potential energy, we can choose a different zero-point, but for elastic potential energy, it is always where the object is at equilibrium.
Consider a block on an ideal spring sliding across a frictionless surface. The energy that is stored as elastic potential energy, \(U_{el}\), in the spring converts to kinetic energy, \(K\), as the block moves. The total mechanical energy of the system, \(E\), is the sum of the elastic potential energy and the kinetic energy at any position, and it is constant in this case since the surface is frictionless. The graph below shows the elastic potential energy of the spring-block system as a function of position. The elastic potential energy is maximized when the spring is at the highest stretched or compressed position, and it is zero when \(x=0\,\mathrm{m}\) at the equilibrium position. The kinetic energy is at the greatest value when the spring is in the equilibrium position, which means that the block's velocity is maximized at that position. The kinetic energy goes to zero at the most stretched and compressed positions.
Total mechanical energy of a block-spring system, Vaia Originals
Elastic Potential Energy Examples
We see examples of elastic potential energy in life every day, such as in trampolines, rubber bands, and bouncy balls. Jumping on a trampoline uses elastic potential energy as the trampoline is stretched when you land on it and pushes you up as you jump again. Springs are used in medical devices, spring mattresses, and numerous other applications. We make use of elastic potential energy from springs in many things that we do!
Elastic potential energy is used when jumping on a trampoline as the springs and material stretch and store energy, Pixabay
A \(0.5\,\mathrm{kg}\) block attached to a spring is stretched to \(x=10\,\mathrm{cm}\). The spring constant is \(k=7.0\,\frac{\mathrm{N}}{\mathrm{m}}\)and the surface is frictionless. What is the elastic potential energy? If the block is released, what is its velocity when it reaches \(x=5\,\mathrm{cm}\)?
We can use the equation for the elastic potential energy of a spring to find the elastic potential energy of the system at \(x=10\,\mathrm{cm}\). The equation gives us:
$$ \begin{aligned} U_{el} &= \frac{1}{2}kx^2\\ &= \frac{1}{2}\left(7.0\,\frac{\mathrm{N}}{\mathrm{m}}\right) \left(0.10\,\mathrm{m}\right) \\ &= 0.035\mathrm{J} \end{aligned}$$
When the block is released, we must also consider the kinetic energy of the system. The total mechanical energy is constant at any position, so the sum of the initial elastic potential energy and the initial kinetic energy is equivalent to their sum when \(x=5\,\mathrm{cm}\). Since the block is not moving initially, the initial kinetic energy is zero. Let \(x_1 = 10\,\mathrm{cm}\) and \(x_2 = 5\,\mathrm{cm}\).
$$ \begin{aligned} K_1 + U_1 &= K_2 + U_2 \\ 0 + \frac{1}{2}kx_1^2 &= \frac{1}{2}mv^2 + \frac{1}{2}kx_2^2 \\ kx_1^2 &= mv^2 + kx_2^2 \\ k\left(x_1^2 - x_2^2\right) &= mv^2 \\ v &= \sqrt{\frac{ k\left(x_1^2 - x_2^2\right)}{m}} \\ v &= \sqrt{\frac{7.0\,\frac{\mathrm{N}}{\mathrm{m}}\left((0.10\,\mathrm{m})^2 - (0.05\,\mathrm{m})^2\right)}{0.5\,\mathrm{kg}}} \\ v &= 0.3\,\frac{\mathrm{m}}{\mathrm{s}} \end{aligned}$$
Thus the velocity at \(x=5\,\mathrm{cm}\) is \(v=0.3\,\frac{\mathrm{m}}{\mathrm{s}}.
Elastic Potential Energy - Key takeaways
- Elastic potential energy is energy that is stored in an elastic object, like a rubber band or a spring, and can be used later.
- The elasticity of an object is how much it can be stretched before going back to its original form.
- The equation for the elastic potential energy of a spring is \(U_{el} = \frac{1}{2}kx^2\).
- The total mechanical energy of a spring-mass system includes kinetic energy and elastic potential energy.
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