1.

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)

Answer»

`(1)/(2)m//s^(2)`
`(3)/(4)m//s^(2)`
`(4)/(3)m//s^(2)`
`(3)/(5)m//s^(2)`

Solution :`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)`


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