Text A spring obeys Hooke’s law F(x)=kxF(x)=kxF(x)=kx with k=200 N/mk=200\ \text{N/m}k=200 N/m. Compute the work needed to stretch it by 0.3 m0.3\text{ m}0.3 m from its rest position. Solution W=∫00.3200x dx=100⋅0.09=9 JW=\displaystyle\int_0^{0.3} 200x\,dx=100\cdot0.09=9\text{ J}W=∫00.3200xdx=100⋅0.09=9 J. W=9 J\boxed{W=9\ \text{J}}W=9 J