Question

An elastic spring of force constant K is compressed by an amount 'x'. Show that its potential energy is (1/2)Kx^2.

Answer 1

We know that as we stretch or compress a spring, a restoring force acts on it which increases linearly with the distance from the equilibrium (unstretched position)

restoring force F is directly proportional to distance stretched/compressed from unstretched position. Let this distance be x

then

F = -kx

k is proportionality constant called 'spring constant'.

(negative sign shows that restoring force is opposite to displacement)

Now we want to know how much potential energy is stored in the compressed spring.

Potential energy of any configuration = work done to get that configuration

So we basically have to calculate the work done by us in compressing the spring by a distance x.

W = Force. Distance

The applied force is equal to the restoring force in magnitude. Let us compress the spring by a small distance dx, then small work done in doing so

dW = kx. dx

Total work done will be equal to integrating all the small works dW from x = 0 to x = x

∫dW = ∫kx. dx

[ from x = 0 to x = x ]

W = k ∫ x.dx = k [x2]/2

W = kx2 /2

Therefore Potential energy stored in the spring = W = kx2 /2

Answer 2

F= kx 

Now as we know 

f.ds = potential energy 

differentiate
 
potential energy = kx .ds 

k is a constant ........ and diff of x .ds is = x² / 2 


so here potential energy will be = 1/2 k x²
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