If a mass of 0.55kg attached to a vertical spring stretches the spring 32.0 cm from its original equilibrium, what is the spring constant?

G weight of mass m=0.55kg.

G=m×g=k×dx, whre dx=32cm is the spring stretch, computed as the difference between the actual length and original length of the spring.

dx=32cm=0.32m

g=9.81m/s²

Thus we have:

m×g=k×dx

k=m×g/dx=0.55×9.81/0.32=16.86N/m

k=16.86N/

f=ma newton's second law 7x8=56n