Question

An unusual spring has a restoring force of magnitude F = (2.0 N/m) x + (1.0 N/m 2) x 2, 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?

Answer 1

Step-by-step explanation:

Given:

Equation

Force, F = (2.00 N/m)x + (1.00 N/m2)x²

Mass, m = 3 kg

Displacement, x = 2.0 m

Workdone, W = F × dx

= integrating f with respect to dx,

F = (2.00 N/m)x + (1.00 N/m2)x²

W = 2x²/2 + x^3/3

At x = 2 m,

Workdone, W = (2)^2 + (2^3)/3

= 4 + 8/3

= 20/3 J

= 6.67 J

Using conservation of energy,

Workdone, W = kinetic energy

= 1/2 × M × v^2

6.67 = 1/2 × 3 × v^2

v = sqrt(4.44)

= 2.108 m/s