elastic potential energy
Summary
Elastic potential energy is energy stored in a deformed elastic object, such as a stretched or compressed spring. For an ideal spring U = ½kx², so it depends on the square of the displacement from equilibrium: doubling the stretch quadruples the energy.
Detail
Meaning: Work done to deform an elastic object is stored as potential energy and can be recovered when the object returns to its natural length. Within the elastic limit a spring obeys Hooke's law, F = −kx. Because the force grows linearly with displacement, the work done is the area under the force versus displacement graph, a triangle, which gives ½kx².
Equation: U = ½kx², where U is elastic potential energy (J), k is the spring constant (N/m), and x is the displacement from the natural, unstretched length (m).
On the MCAT: Expect a mass on a spring or a block pushed against a compressed spring. With no friction, mechanical energy is conserved: ½kx² + ½mv² stays constant. At maximum displacement (the amplitude) all the energy is elastic; at the equilibrium position it is all kinetic. A launch problem sets ½kx² = ½mv² to find speed. You may also be asked for the work from the area under an F versus x graph, or to predict that doubling x quadruples U. For a vertical spring, gravitational potential energy must be included too.
Common confusion: x is the change from natural length, not the spring's total length. Stretching and compressing by the same amount store the same energy, since x is squared. Doubling the stretch doubles the force but quadruples the energy.
Sources
- Physics
- Fundamentals of Physics
- University Physics
Written for the MCAT and checked before it was published. For study reference only. Spotted an error? Let us know.