Question

An object is attached to a spring of spring constant 42 N/m and has 700 J of elastic potential
energy. How far does it extend?

Use Formula: PEelastic = KX2 / 2 K is spring contstant (n/m) , x is distance (m)

Answer 1

Use PE = 1/2 k x^2. Solve for x:

x^2 = 2 PE / k = 2(700) / 42 = 1400 / 42 = 100/3

x = sqrt(100/3) = 10 / sqrt(3) ≈ 5.77 m.

So the spring extends about 5.77 m.