Answer:
Elastic potential energy, E = 2 J
Explanation:
It is given that,
Spring constant of the spring, k = 2500 N/m
The spring is stretched to a distance of 4 cm i.e. x = 0.04 m
We have to find the elastic potential energy possessed by the spring. A spring possessed elastic potential energy and it is given by:
[tex]E=\dfrac{1}{2}kx^2[/tex]
[tex]E=\dfrac{1}{2}\times 2500\ N/m\times (0.04\ m)^2[/tex]
E = 2 Joules.
Hence, the correct option is (d) " 2 Joules ".
