Elastic Potential Energy - Learn

Elastic Potential Energy

Hooke’s Law is used to determine the force applied to a spring when it is extended.

Hooke’s Law equation:

$F=-kx$

where;

$F=$ force applied to the spring in $N$

$k=$ the spring constant in $Nm^{ -1 }$

$x=$ the extension of the spring in $m$

Note that the minus sign in the equation refers to the restoring force within the spring that is opposing the applied force that caused the extension.

When a spring is stretched or compressed, it stores energy. This energy is known as elastic potential energy. The potential energy stored in a spring is given by:

$U_{ p }=\cfrac { 1 }{ 2 } k\Delta x^{ 2 }$

where;

$U_{ p }=$ potential energy stored in a spring in $J$

$k=$ the spring constant in $Nm^{ -1 }$

$\Delta x=$ the extension of the spring in $m$

Applying the law of conservation of energy, $U_{ p }$ is also equal to:

The work done by the force causing the extension, or the compression.
The work done by the spring compressing or expanding back to its original length.
The work done on any projectile which might be fired by the compressed spring.
The kinetic energy gained by this fired object.

Example 1:

A spring has a spring constant of 25 $Nm^{ -1 }$ . What force is required to extend it by 10cm?

Using $F=-kx$

$F=-25\times 0.1$ (10cm = 0.1m)

$F=-2.5N$

Applied force, $F=2.5N$

Example 2:

A spring has a spring constant of 60 $Nm^{ -1 }$ . How far does it extend when a force of 15N is applied to it?

Using $F=-kx$

$15=-60\times x$

$x=-\cfrac { 15 }{ 60 }$

$x=0.25m$

The spring would extend 25cm

Example 3:

A projectile is launched by a spring with a spring constant of 400 $Nm^{ -1 }$ . Calculate the potential energy of the projectile if the spring was compressed 20cm:

$U_{ p }=\cfrac { 1 }{ 2 } k\Delta x^{ 2 }$

$U_{ p }=\cfrac { 1 }{ 2 }\times 400\times 0.2^{ 2 }$

$U_{ p }=8J$

Example 4:

A catapult launches a rock with a spring mechanism that has a spring constant of 650 $Nm^{ -1 }$ . How far would the spring need to be extended for the rock to gain 30J of energy?

$U_{ p }=\cfrac { 1 }{ 2 } k\Delta x^{ 2 }$

$30=\cfrac { 1 }{ 2 }\times 650\Delta x^{ 2 }$

$x^{ 2 }=\cfrac { 30 }{ 325 }$

$x^{ 2 }=0.092$

$x=0.30$

The spring would need to be compressed 30cm