Two blocks of masses 8 kg and 12 kg are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system is

Question

Two blocks of masses 8 kg and 12 kg are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system is (A) (g/4) (B) (g/5) (C) (g/8) (D) (g/6)

Tardigrade Admin · Accepted Answer

Let a be the common acceleration of the system and T be tension of the string. The equations of motion of two blocks are T-8g=8a ldots(i) and 12g-T=12a ldots(ii)<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/dd0a414979a3d44e884644b0d1a69ec6-.png" /> Adding (i) and (ii), we get 4g=20a or a=(g/5)

Q. Two blocks of masses $8 k g$ and $12 k g$ are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system is

$\frac{g}{4}$

$\frac{g}{5}$

$\frac{g}{8}$

$\frac{g}{6}$

Let $a$ be the common acceleration of the system and $T$ be tension of the string. The equations of motion of two blocks are
$T - 8 g = 8 a$ $\dots (i)$
and $12 g - T = 12 a$ $\dots (ii)$
Adding (i) and (ii), we get
$4 g = 20 a$ or $a = \frac{g}{5}$

Q. Two blocks of masses 8kg and 12kg are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system is

4g​

5g​

8g​

6g​