Q. Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force $F$ applied on the upper string produces an acceleration of $2\ m/s^{2}$ in the upward direction in both the blocks. If $T$ and $T^{'}$ be the tensions in the two parts of the string, then $(\ g = 9.8\ m/s^{2})$
A) $T = 70.8\ N$ and $T^{'} = 47.2\ N$
B) $T = 58.8\ N$ and $T^{'} = 47.2\ N$
C) $T = 70.8\ N$ and $T^{'} = 58.8\ N$
D) $T = 70.8\ N$ and $T^{'} = 0$
Explanation
(a) FBD of mass $2\ kg\ FBD$ of mass $4kg$
$TT^{'} - 19.6 = 4$ $\ldots(i)$
$T^{'} - 39.2 = 8$ $\ldots(ii)$
From (ii), $T^{'} = 47.2\ N$
And substituting $T^{'}$ in (i), we get
$T = 4 + 19.6 + 47.2 \Rightarrow T = 70.8\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.