Q.
Two blocks of masses $M$ and $m$ are connected by a string passing over a pulley as shown in
the figure. The downward acceleration of the block with mass $m$ is
A) $M/(m + M)g$
B) $mg/(m + M)$
C) $(m + M)/mg$
D) $(n + M)/Mg$
Explanation
(a) To blocks of masses M and $m\ $are connected by a string passing over a pulley. Let $T\ $is the tension in the string and $a$ is the downward acceleration.
Applying II law of motion for block of mass M.
$T = ma$ …(i)
For block of mass $m$
$ma = mg - T$ …(ii)
[image:5_image157.jpeg]
From Eqs. (i) and (ii), we get
$ma = mg - Ma$
$or\ \ a = \frac{mg}{m + M}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.