5. A certain spring that obeys Hooke's Law stretches 30 cm when a load of 0.35 N is added to it. How much energy is stored in the spring when it is compressed 5.0 cm?

Energy stored in the spring when compressed 5.0 cm = 0.00146 J = 1.46 mJ

Explanation:

According to Hookes law,

F = kx

where F = force or load applied = 0.35 N

k = spring's constant = ?

x = extension or compression of the spring = 30 cm = 0.3 m

0.35 = k * 0.3

k = 0.35/0.3 = 1.167 N/m

Energy stored in a spring = kx²/2

k = 1.167 N/m, x = 5 cm = 0.05 m

Energy stored in the spring = (1.167) (0.05²) / 2 = 0.00146 J = 1.46 mJ