Question

Three blocks are connected by massless strings on a frictionless inclined plane of $30^{\circ}$ 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, $\left.g=10\, m / s ^{2}\right)$

• 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 1

Answer: (D)

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}-8 \,g \,\sin 30^{\circ}=8 a $
$\Rightarrow 104-T_{1}-8 \times 10 \times \frac{1}{2}=8 a \,\,\left[\because g=10\, m / s ^{2}\right] $
$64-T_{1}=8 a \ldots( i )$
Equation of motion for $5 \,kg$ block is
$T_{1}-T_{2}-5\, g \,\sin 30^{\circ} =5\, a $
$T_{1}-T_{2}-25 =5\, a\,\,\,...(ii)$
Equation of motion for $3 \,kg$ block is
$T_{2}-3 \,g \,\sin 30^{\circ} =3 a$
$T_{2}-15 =3 a\,\,\,...(iii)$
Adding Eqs. (i), (ii) and (iii), we get
$24=16 a $
$\Rightarrow a=1.5\, m / s ^{2}$
Hence, acceleration $a$ of block is $1.5 \,m / s ^{2}$.