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

 

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