physics
posted by Kaylee .
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

Using the equation for spring potential energy
W=(1/2)kx²
so
ΔW=(1/2)k(x2²x1²)
=(1/2)(30 N/m)*(0.96²0.19²)
=13.3 J
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