A block of mass `m=4kg` is attached to a spring of spring constant `(k=32 Nm^(-1))` by a rope that hangs over a pulley of mass `M=8kg` If the system starts from rest with the spring unstretched, find the speed of the block after it falls `1m`. Treat the pulley as a disc, so `I=1/2 MR^2`

Explanation

Since the rim of the pullely moves at the same speed as the block, the speed of the block and the angular velocity of the pulley related by `v= omegaR` when the block falls by a distance `x`, its potential energy decrease `(DeltaU_(g)=-mgx)`, the potential energy of the spring increases `(Delta_(s)=+1/2kx^(2)`), and both the block and the pulley gain `KE`
`(DeltaK=1/2mv^(2)+1/2Iomega^(2))`
From the conservation of mechanical energy,
`DeltaK+DeltaU=0`,
`1/2mv^(2)+1/2I(v/R)^(2)+1/2kx^(2)-mgx=0`
`1/2(m+M/2)v^(2)+1/2kx^(2)-mgx=0`
Putting `m=4kg, M=8 kg, k=32 Nm^(-1), x=1 m`
`1/2(4+8/2)v^(2)+1/2(32)(1)^(2)-(4)(10)(1)=0`
`4v^(2)+16-40=0 rArrv=2.4 mx^(-1)`