Question

A block of mass `m=5kg` is placed on the wedge of mass `M=32kg` as shown in the figure. Find the acceleration of wedge with respect to ground. (Neglect any type of friction. Spring and pulley are ideal)

A. `(1)/(2)m//s^(2)`
B. `(3)/(4)m//s^(2)`
C. `(4)/(3)m//s^(2)`
D. `(3)/(5)m//s^(2)`

Answer 1

`T(1+costheta)-Nsintehta)=Ma`……(`i`)
FBD of wedge from ground frame
`mg sintheta-ma costheta-T=ma`
`rArr mg sintheta-T=ma(1+costheta)`……(`ii`)
`N=m(gcostheta+asintheta)`…….(`iii`)
Using `(i) +(ii) (1+costheta)+ (iii) sintheta`
`mg sintheta+ mg sintheta costheta=`
`Ma+ma(1+costheta)^(2)+mg sintheta costheta+masin^(2)theta`
`rArr a=(mg sintheta)/(M+2m(1+costheta))`
given `theta=37^(@)`, `m=5kg` and `M=32kg`
so, `a=(3)/(5)m//s^(2)`