What is the potential energy of a 0.4 m spring when stretched to 1/4 of its length? Spring stiffness coefficient 300 N / m.

These tasks: l (length of the stretched spring) = 0.4 m; Δl (tensile strain) = 1/4 l = 0.25l; k (stiffness factor) = 300 N / m.

The potential energy acquired by the considered spring under tension is determined by the formula: En = k * Δl ^ 2/2.

Let’s make a calculation: En = 300 * (0.25 * 0.4) ^ 2/2 = 1.5 J.

Answer: The considered spring, when stretched to 1/4 of its length, will acquire a potential energy of 1.5 J.