Three blocks are connected by massless strings on a frictionless inclined plane of `30^(@)` as shown in the figure. A force of 104 N is applied upward along the incline to mass `m_(3)` causing an upward motion of the blocks. What is the acceleration of the blocks? (Assume, acceleration due to gravity, `g=10 m//s^(2)`)

Options

A. `6.0 m//s^(2)`
B. `4.5 m//s^(2)`
C. `3.0 m//s^(2)`
D. `1.5 m//s^(2)`

Answer

D

Explanation

The given situation is shown in the figure below,

If the system moves upward with acceleration a `m//s^(2)` on the application of 104 N, then the equation of motion for 8 kg block is
`104-T_(1)-8g sin 30^(@)=8a`
`implies 104-T_(1)-8xx10xx1/2=8a" "[ :' g=10 m//s^(2)]`
`64-T_(1)=8a` ...(i)
Equation of motion for 5 kg block is
`T_(1)-T_(2)-5 g sin 30^(@)=5 a`
`T_(1)-T_(2)-25=5 a` ...(ii)
Equation of motion for 3 kg block is
`T_(2)-3g sin 30^(@)=3a`
`T_(2)-15=3a` ...(iii)
Adding Eqs. (i), (ii) and (iii), we get
`24= 16 a implies a=1.5 m//s^(2)`
Hence, acceleration a of block is `1.5 m//s^(2)`.