Answer to Question #174243 in Physics for Quincy

Question

A spring has force constant of 10 N/m. How much work is done to extend it 10 mm from equilibrium position? How much extra work is needed to extend an additional 10 mm? Are they the same?

Answer 1

We can find the work done to extend string 10 mm from equilibrium position as follows:

$W_1=\dfrac{1}{2}kx^2=\dfrac{1}{2}\cdot10\ \dfrac{N}{m}\cdot(10\cdot10^{-3}\ m)^2=5\cdot10^{-4}\ J.$

Let's find the the work done to extend string 20 mm from equilibrium position:

$W_2=\dfrac{1}{2}kx^2=\dfrac{1}{2}\cdot10\ \dfrac{N}{m}\cdot(20\cdot10^{-3}\ m)^2=0.002\ J.$

Therefore, we can find the extra work to extend spring an additional 10 mm:

$\Delta W=W_2-W_1=0.002\ J-5\cdot10^{-4}\ J=0.0015\ J.$

As we can see they aren't the same.