a spring initially has a potential energy of 20J. Then, a force is applied on the spring and causes it to stretch by half its length. How much potential energy does the spring now have? the force constant of the spring is 200N/m
Answer (1)
Answer:The potential energy of a spring is given by the formula:PE = (1/2) * k * x^2Where: * PE is the potential energy * k is the force constant * x is the displacement from the equilibrium positionIn this case, the force constant (k) is given as 200 N/m. We are also told that the spring stretches by half its length. Let's assume the initial length of the spring is L. So, the displacement (x) is L/2.Initially, the potential energy is 20 J. We can use this information to find the initial displacement (x0):20 = (1/2) * 200 * x0^2x0^2 = 0.2x0 = sqrt(0.2)Now, we can find the final potential energy using the new displacement (x = L/2):PE_final = (1/2) * 200 * (L/2)^2PE_final = (1/2) * 200 * (1/4) * L^2PE_final = (1/8) * 200 * L^2Since we know that the initial displacement is sqrt(0.2), we can substitute this value into the equation for PE_final:PE_final = (1/8) * 200 * (sqrt(0.2))^2PE_final = (1/8) * 200 * 0.2PE_final = 5 JTherefore, the spring now has a potential energy of 5 J.
