Question

4.0 J of work are performed in stretching a spring with a spring constant of 2500 N/m. How much is the spring stretched?

Answer 1

$energy = elastic \: potential$

$4 = \frac{1}{2} kx {}^{2}$

$4 = \frac{1}{2} (2500)x {}^{2}$

$4 = 1250x {}^{2}$

$x {}^{2} = \frac{4}{1250} = 0.0032 \: meters$

$x = \sqrt{0.0032} = 0.05657 \: meters$