Question

In the figure shown, find the acceleration (in $m/s^2$) of the $2kg$ body and the tension (in $N$) in the string. (String is massless and inextensible)

• A. $0,2g$
• B. $g,2g$
• C. $2g,g$
• D. $0,g$

Answer 1

The correct option is A $0,2g$

FBDs of both the masses are as shown

Let the direction of the acceleration $a$, of the systems be as shown in the figure and the tension in the string be $T$

From the FBDs we have
$T-4g.sin{30}^{\circ }=4a⟹T-2g=4a$ ............(i)
And, $2g-T=2a$..........(ii)
From eq (i) & (ii) we have
$a=0$
Also, $2g-T=0⟹T=2g$

Alternate:
If both the bodies are considered as a system then acceleration of the system is given by
$a=\frac{\text{(Supporting Force)-(opposing Force)}}{\text{Total Mass}}=\frac{2g-4g.sin{30}^{\circ }}{6}=0$
Also, from the FBD of $2Kg$
$2g-T=2a⟹T=2g$