Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :

Question

Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is : (A) g (B) g/2 (C) g/5 (D) g/10

Tardigrade Admin · Accepted Answer

The acceleration of the system a=((m2-m1)g/m1+m2) where m2> m1 a=((6-4)g/6+4) =(2g/10) a=(g/5)

Q. Two bodies of mass $4 k g$ and $6 k g$ are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :

g

g/2

g/5

g/10

The acceleration of the system $a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}}$ where $m_{2} > m_{1}$
$a = \frac{( 6 - 4 ) g}{6 + 4}$
$= \frac{2 g}{10}$
$a = \frac{g}{5}$

Q. Two bodies of mass 4kg and 6kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :

g

g/2

g/5

g/10

The acceleration of the system a=m1​+m2​(m2​−m1​)g​ where m2​>m1​ a=6+4(6−4)g​ =102g​ a=5g​

Q. Two bodies of mass $4 k g$ and $6 k g$ are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :

The acceleration of the system $a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}}$ where $m_{2} > m_{1}$
$a = \frac{( 6 - 4 ) g}{6 + 4}$
$= \frac{2 g}{10}$
$a = \frac{g}{5}$