55.8 cm
Using Hooke's law, the energy stored in a spring is
E = 0.5 kx^2
where
E = Energy
k = spring constant
x = deflection of spring
So substituting the known values and solving for x, we get:
E = 0.5 kx^2
280 J = 0.5 1800 N/m x^2
280 N*m = 900 N/m x^2
0.311111111 m^2 = x^2
0.557773351 m = x
Rounding to 3 significant figures, gives us 55.8 cm.
So the spring must be stretched 55.8 cm to store 280 Joules of energy.
