Question

A spring with a spring constant of 50 n/m is attached to a 12 kg mass. From its equilibrium point (unstretched spring) the block is pulled back a distance of .30 m The block is let go and it begins to move back to its equilibrium point. What is the velocity of the block when it reaches the equilibrium point? (Assume no friction.)

Answer 1

Given,. Mass m = 12 kg

Spring constant k = 50 N/m

Compression in spring x = 0.30 m

At equilibrium position, the potential energy of spring will become equal to the kinetic energy of the block

Hence, 1/2 kx^{​​​​​2}​ = 1/2 mv^​​​v2

Or, kx^​​​2 = mv^{​​​​​2}

Or, v^{​​​​​2} = kx^​​​2 /m = 50×0.3² / 12 = 50× 0.09/ 12 = 0.375

Or, v = (0.375)^1/2 = 0.61 m/s

Hence, velocity at equilibrium poinnt is 0.61 m/s .