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 25Nm^{ -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 60Nm^{ -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 400Nm^{ -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 650Nm^{ -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