Question

A cylindrical spring is found to elongate by 3 cm when stretched by a force of 500N.

Determine: 

 a) its force constant

 b) the work done by a 600N force in elongating the spring

Answer 1

a) $F=k\Delta x\to k=F/\Delta x=500/0.03=16667\ (N/m)$

b) $W=PE=\frac{k(\Delta x)^2}{2}=\frac{k(F/k)^2}{2}=\frac{16667\cdot(600/16667)^2}{2}=10.8\ (J)$