A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the mass is pulled down to where the spring has a length of 1.01 m and given an initial speed upwards of 1.6 m/s. What is the maximum length of the spring during the motion that follows?

Answer;

= 1.256 m

Explanation;

We can start by finding the spring constant

F = k*y

Therefore; k = F/y = m*g/y

= 0.40kg*9.8m/s^2 / (0.76 - 0.41)

= 11.2 N/m

Energy is conserved

Let A be the maximum displacement

Therefore; 1/2*k*A^2 = 1/2*k * (1.20 - 0.41) ^2 + 1/2*m*v^2

Thus; A = sqrt ((1.20 - 0.55) ^2 + m/k*v^2)

= sqrt ((1.20 - 0.55) ^2 + 0.40/9.8*1.6^2)

= 0.846 m

Thus; the length will be 0.41 + 0.846 = 1.256 m