A 35.0 kg mass is suspended at equilibrium on a spring that has a spring constant of 1.23*10^3 N/m. The mass is pulled down by 18.5 cm...from its...

The spring has a spring constant of 1.23*10^3 N/m. A 35 kg
mass is suspended from it. The mass is pulled down so that the spring is extended by
18.5 cm. It is then released and the rises up by 27
cm.

After the mass has risen 27 cm it is at a height equal
to 27 - 18.5 = 8.5 cm above the equilibrium position of the spring. There are two forces
acting on the mass here:

• A force due to the
  gravitational force of attraction of the Earth which accelerates it downwards by 9.8
  m/s^2

• A force exerted by the spring
  downwards which is equal to 1.23*10^3*8.5*10^-2 = 104.55 N. This accelerates the spring
  downwards by 104.55/35 = 2.987 m/s^2

The net
acceleration of the body is 9.8 + 2.987 = 12.787
m/s^2.

At the required position the
acceleration of the body is 12.787 m/s^2
downwards.