An unusual spring has a restoring force of magnitude F 20 Nm

An unusual spring has a restoring force of magnitude F =(2.0 N/m) x + (1.0 N/m^2) x^9 where x is the stretch of the spring from its equilibrium length. A 3.00 kg mass is attached to this spring and released from rest after stretching the spring 2.00 m. What is the speed of the mass when the spring returns to its equilibrium length?

Solution

Here,

F = 2 x + 1 x^2

x = 2 m

m = 3 Kg

let the speed of mass is v

Using work energy theorum

kinetic energy of block = work done by spring

0.5 * 3 * v^2 = (F * dx) from 0 to 2 m

1.5 * v^2 = ((2 x + 1 x^2 ).dx) from 0 to 2 m

1.5 * v^2 = (x^2 + x^3/3) from 0 to 2 m

1.5 * v^2 = (2^2 + 2^3/3)

solving for v

v = 2.108 m/s

the speed of mass m when it returns to equilibrium length is 2.108 m/s