With a deformation of 2 cm, the steel spring has a potential energy of elastic deformation of 4 J. How much will the potential

With a deformation of 2 cm, the steel spring has a potential energy of elastic deformation of 4 J. How much will the potential energy of this spring change with a decrease in deformation by 1 cm?

Spring potential energy:
Ep1 = (k * ∆l1²) / 2, where A is the potential energy of the spring (Ep = 4 J), k is the stiffness of the spring (N / m), ∆l1 is the amount of spring deformation (∆l1 = 2 cm = 0.02 m).
Let us express and calculate the stiffness of the spring:
k = 2Ep / ∆l1² = 2 * 4 / 0.02² = 20,000 N / m.
Potential energy with a decrease in deformation by 1 cm (∆l2 = 2 – 1 = 1 cm = 0.01 m):
En2 = (k * ∆l2²) / 2 = 20,000 * 0.01² / 2 = 1 J.
Potential energy change:
En1 – En2 = 4 – 1 = 3 J.
Answer: Potential energy will decrease by 3 J.