Amass 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?

= 1.256 m

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

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr