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

Smart Achievers · Accepted Answer

(2)

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 $12 g - T = 12 a$ $\dots (i i)$

Adding (i) and (ii), we get

$4 g = 20 a$ or $a = \frac{g}{5}$

Solution

Q.
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

1 $\frac{g}{4}$

2 $\frac{g}{5}$

3 $\frac{g}{8}$

4 $\frac{g}{6}$

Ans.
(2)

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 $12 g - T = 12 a$ $\dots (i i)$

Adding (i) and (ii), we get

$4 g = 20 a$ or $a = \frac{g}{5}$

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Solution

Q. 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

1 g4

2 g5

3 g8

4 g6

Ans. (2) 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 ⋯(i) and 12g-T=12a ⋯(ii) Adding (i) and (ii), we get 4g=20a or a=g5