Question

The force to pull a nonlinear spring is function of the stretched distance x (the displacement of the spring from its unstretched length) measured in meters. If F-292x3 + 30x2 (N) Find the work needed to compress the spring a distance of 0.55m.

Answer 1

Answer: work = 8.34 Nm

Hi! The work W done by a force F to move an object from a to b is defined as (in the one dimensional case):

$W = \int\limits^a_b {F}\, dx$  

In this case

$F = 292x^3 + 30x^2$

$W = \int\limits^{0.55}_0 ({292x^3 + 30*x^2}) \, dx$

$W = \frac{292}{4} x^4 + \frac{30}{3} x^3 \Big|_{0}^{0.55} = 8.34$