Question

A 0.50 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 325 N/m) whose other end is fixed. The mass has a kinetic energy of 16.0 J as it passes through its equilibrium position (the point at which the spring force is zero). At what rate is the spring doing work on the mass as the mass passes through its equilibrium position?

At what rate is the spring doing work on the mass when the spring is compressed 0.157 m and the mass is moving away from the equilibrium position?

Answer 1

part a )

0 watt because at equilibrium, there is no force from the spring acting on the mass. As work is Force times distance, with no force, no work is done and no power is applied as power is the rate of doing work.

part b )

At 0.157 m

U = 1/2*k*x^2

U = 1/2 * 325N/m * (0.157m)^2 = 4.005 J

at equilibrium point total energy = 1/2*mv^2 + 1/2*kx^2

KE = 16J - 4.005J = 11.995 J

KE = 1/2mv^2 = 1/2 * 0.5kg * v^2

11.995 = 1/2 * 0.5 * v^2

v = 6.93 m/s

F = kx = 325N/m * 0.157m = 51.025 N

rate of work = power = F * v = 51.025 * 6.93m/s = 353.60 W