A spring with spring constant of 30 N/m is stretched 0.19 m from its equilibrium position. How much work must be done to stretch it

an additional 0.077 m? Answer in units of J

1 answer

Using the equation for spring potential energy
W=(1/2)kx²

so
ΔW=(1/2)k(x2²-x1²)
=(1/2)(30 N/m)*(0.96²-0.19²)
=13.3 J