A spring with a stiffness of 103 N / m is stretched by 4 cm. What is the potential energy of elastic deformation of the spring?
| July 22, 2021 | education
Initial data: k (spring rate) = 103 N / m; Δl (amount of tension of the spring) = 4 cm.
SI system: Δl = 4 cm = 0.04 m.
The potential energy of elastic deformation of a spring with given parameters is determined by the formula: En = k * Δl ^ 2/2.
Let’s perform the calculation: Ep = 103 * 0.04 ^ 2/2 = 0.8 J.
Answer: The potential energy of elastic deformation of the spring is 0.8 J.
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