How far must you stretch a spring with k = 1800 n/m to store 280 j of energy?

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.

Bag of bolts weight 29.4n