The correct answer to the question is 20.07 J.
CALCULATION:
The mass of the block attached to the spring is given as m = 25 kg.
The force or spring constant of the spring is given as k = 95 N/m.
Due to the block, the spring was compressed to a distance x = 0.65 m.
We are asked to calculate the potential energy of the spring. When the spring is compressed, there will be change in its configuration . Due to this change, the spring will gain elastic potential energy.
The elastic potential energy of the spring is calculated as -
Potential energy P.E = [tex]\frac{1}{2}kx^2[/tex]
= [tex]\frac{1}{2}\times 95\ N/m \times (0.65m)^2[/tex]
= 20.07 Joule [ans]
Hence, the potential energy of the spring is 20.07 J.
