Question

A spring with spring constant of 21 N/m is
stretched 0.13mfromits equilibriumposition.
How much work must be done to stretch it
an additional 0.12 m?
Answer in units of J

Answer 1

\[W=\Delta U=\frac{1}{2}k (\Delta x)^2=\frac{1}{2}\cdot21\cdot0.12^2=0.151\ J\]

Answer 2

It said that .151 was incorrect it there another way to do it?

Answer 3

Yes, I think I missunderstood the problem. It should be that way,
\[W=\Delta U=\frac{1}{2}kx_f^2-\frac{1}{2}kx^2_0=\frac{1}{2}\cdot21\cdot(0.25^2-0.13^2)=0.479\ J\]Where 0.25=0.13+0.12 is the final deformation.

Answer 4

Thank you :)

Answer 5

;)