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Question: An elastic spring of unstretched length L and force constant K is stretched by a small amount x. It ...

An elastic spring of unstretched length L and force constant K is stretched by a small amount x. It is further stretched by small length y. The work done in the second stretching is
A.12Ky2\dfrac{1}{2}K{{y}^{2}}
B.12K(x2+y2)\dfrac{1}{2}K({{x}^{2}}+{{y}^{2}})
C.12Ky(2x+y)\dfrac{1}{2}Ky(2x+y)
D.12K(x+y)2\dfrac{1}{2}K{{(x+y)}^{2}}

Explanation

Solution

The total work done in stretching wire by a length l is given by,
W=12×load×extensionW=\dfrac{1}{2}\times load\times extension. This total work in stretching the wire gets stored in the form of its elastic potential energy. It is sometimes called strain energy.

Complete answer:
We have an elastic string having L. Force constant is K. Now this is stretched by a small amount x. Again the same spring is stretched by y after stretching x. So the final length of spring is (L+x+y). We need to find out the work done in stretching y length.
Aim: The work done in the second stretching is?
We know that work done is nothing but the amount of force acting on the body to displace it by a length. In simple words work done is a product of force and displacement. In this case, displacement is the final distance that is x+y.
So final length is x+y
Initial length is x
The final length is y
So in initial case work done in displacing or stretching length x is given by,
Winitial=12Kx2{{W}_{initial}}=\dfrac{1}{2}K{{x}^{2}}
In final case work done in displacing or stretching length (x+y) is given by,
Wfinal=12K(x+y)2{{W}_{final}}=\dfrac{1}{2}K{{(x+y)}^{2}}
Now we need to find out work done in stretching y length is deNoted by Wy{{W}_{y}}
To calculate the second length:
Wy=WfinalWinitial=12K(x+y)212Kx2{{W}_{y}}={{W}_{final}}-{{W}_{initial}}=\dfrac{1}{2}K{{(x+y)}^{2}}-\dfrac{1}{2}K{{x}^{2}}
Wy=12K(x2+y2+2xy)12Kx2{{W}_{y}}=\dfrac{1}{2}K({{x}^{2}}+{{y}^{2}}+2xy)-\dfrac{1}{2}K{{x}^{2}}
Wy=12Ky(2x+y){{W}_{y}}=\dfrac{1}{2}Ky(2x+y)
Work done in stretching y length is Wy=12Ky(2x+y){{W}_{y}}=\dfrac{1}{2}Ky(2x+y)

The correct answer is option (D).

Note:
We know that whenever there is a case of spring, there is force acting known as force constant. Final work down is equal to initial work down plus final work down. Work done is dependent on force constant and length elongation.