Integrate `3x^2 (cosx^3+8)`.
`a) sinx^3−8x^3+C`
`b) sinx^3+8x^3+C`
`c) –sinx^3+8x^3+C`
`d) sinx^3−x^3+C`

Explanation

Suppose the deceleration of the block is a. The linear decelerationof the rim of the wheel is also a. The angular decelratinof the wheel is `alpha=a/r`. I the tension in the string is T, the equations of motion are as follows:
`Mgsintheta-T=Ma`
and `Tr=Ialpha=Ia/r`
ElimiN/Ating T from these equations
`Mgsintheta-Ia/r^2=Ma`
giving `a=(Mgr^2sintheta)/(I+Mr^2)`
THe initial velocity of the block up the incline is v=omegar`. Thus the distance moved by the block before stopping is
`x=v^2/(2a) =(omega^2r^2(I+Mr^2))/(2Mgr^2sintheta)=((I+Mr^2)omega^2)/(2Mgsintheta)`