Answer:
16 J
Explanation:
It is given that,
Work done, W = 2 J
A spring is stretched by 2.0 cm from its equilibrium length
We need to find how much more work will be required to stretch it an additional 4.0 cm.
Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :
[tex]U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m[/tex]
The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J[/tex]
So, the required work done is 16 J.
