My attempt to solve the first one.
T-mg=ma
T=m(g+a)
TR = Iα
I=(1/2)(M)(R^2)
α=a/R
TR=(1/2)(M)(R^2)(a/R)
T=(1/2)Ma
mg+ma=1/2Ma
a=(2mg)/(M-2m)
And... I got a wrong answer. As usual.
According to the first equation, the final potential energy is equal to the initial kinetic energy of the block. So that means that Vblok is the instantaneous speed of the block right before it moves to the right and compress the spring, right? But doesn't the second equation (The initial total...