In experiment 1, a block of mass m is attached to the end of vertical spring of spring constant k0 with its free end at vertical position l0, as shown in figure 1. The mass of the spring is considered to be negligible. When the block is attached to the spring and is at rest at the block-spring’s equilibrium position, the spring is stretched so that its end is at a new position l1, as shown in figure 2. The block is then pulled down to a new vertical position l2 and then released from rest so that the block-spring system oscillates. Assume that the reference line for zero gravitational potential energy of the system is at the lowest point in the system’s vertical displacement from equilibrium. The experiment is assumed to be performed near earth’s surface. What is the magnitude of the change in potential energy of the block-spring system when it travels from its lowest vertical position to its highest vertical position?.

The magnitude of the change in potential energy of the system when it travels from its lowest vertical position L₂ to its highest vertical position L₁ is ΔPE = 2KL₁L₂.

What is the elastic potential energy, PE?

The elastic potential energy is the one that accumulates an elastic body that can be deformed -spring-.

The force that is applied to a spring is directly proportional to deformation. This leads to the following formula to calculate the force,

F = K X = K ΔL = K (L- L₀)

Where

F is the force ⇒ Newton
K is the elastic constant that depends on the spring material ⇒ N/m
X is the deformation ⇒ m
L₀ is the initial position ⇒ m
L is the final position ⇒ m

This relationship between terms is known as Hook's Law.

The energy that the spring accumulates when it is stretched (or comprised) is the elastic potential energy.

This energy is what makes the spring tend to go back to its original position, and can be calculated by using the following formula,

PE = 1/2 K X² = 1/2 K(ΔL)² = 1/2 K (L-L₀)²

This expression tells us which is the elastic potential energy accumulated in a spring that has been deformed. PE is always positive (or zero) and it is expressed in Joules (= Nm).

In the exposed example, we have three positions,

L₀ ⇒ When the spring is at position L₀ ⇒ X = 0 and PE = 0

L₁ ⇒ When the spring is at position L₁ ⇒ X = 0 and PE = 1/2 K (L₁-L₀)²

L₂ ⇒ When the spring is at position L₂ ⇒ X = 0 and PE = 1/2 K (L₂-L₀)²

The magnitude of the change in potential energy of the system when it travels from its lowest vertical position to its highest vertical position is

ΔPE = 2KL₁L₂

You can learn more about the elastic potential energy at

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