Question

An unusual spring has a restoring force of magnitude F = (2.00 N/m)x + (1.00 N/m2)x2, where x is the stretch of the spring from its equilibrium length. A 3.00-kg object is attached to this spring and released from rest after stretching the spring 1.50 m. If the object slides over a frictionless horizontal surface, how fast is it moving when the spring returns to its equilibrium length?

Answer 1

given,

mass = 3 kg

x = 1.50 m

Restoring Force

F = (2.00 N/m)x + (1.00 N/m2)x²

work done =
`\int F.dx`

=
`\int( (2.00)x + (1.00)x^2)dx`

=
`x^2 +(1)/(3)x^3`

now work done at x = 1.5

=
`x^2 +(1)/(3)x^3`

=
`1.5^2 +(1)/(3)1.5^3`

W= 3.375 J

we know

`W = (1)/(2)mv^2`

`3.375 = (1)/(2)* 3 * v^2`

v = 1.5 m/s