Find the magnitude of acceleration (in ${m/s}^{2}$ ) of the system if all surfaces are smooth. $g = 10 {m/s}^{2}$ .

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Solution

The correct option is B 3
Acceleration of the blocks are shown in the FBD

So, from the FBDs we have
$T_{2} - 4 g . s i n 30^{\circ} = 4 a$ .........(i)
$5 g - T_{1} = 5 a$ ...................(ii)
On solving (i) & (ii) we have
$T_{2} - T_{1} + 3 g = 9 a$ ............(iii)
Now, from the FBD of $1 K g$ mass
$T_{1} - T_{2} = a$ ...........(iv)

From eq (iii) & (iv) we have,
$3 g = 10 a ⟹ a = \frac{30}{10} = 3 m / s^{2}$

Alternate:

Supporting force = $5 g = 5 \times 10 = 50 N$
Opposing force = $4 g . s i n 30^{\circ} = 2 g = 20 N$
Magnitude of acceleration of the system is given by,
$a = \frac{Supporting Force - Opposing Force}{Total mass}$
$= \frac{5 g - 2 g}{10} = 3 {m/s}^{2}$

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