Question

6.

Two bodies of masses 8 kg and 4 kg are connected to ends of a string that passes
over smooth pulley if the system is released from rest then acceleration each
mass is
​

Answer 1

Answer:

Explanation:Let the tension be T and acceleration be a. Then,

From  FBD of block of mass 8Kg  

T=8a               ..... eq.1

And, From  FBD of block of mass 2Kg  

2g−T=2a

⇒T=20−2a=8a                  from eq.1

⇒a=2m/s  

2

 

⇒T=8a=16N  

Answer 2

Let two bodies of masses $\sf{m_1}$ and $\sf{m_2}$ each be connected to ends of a massless and inextensible string that passes over a smooth pulley. Let $\sf{m_1>m_2.}$

The acceleration of each body will be the same in magnitude since the string is inextensible. Let it be $\sf{a.}$

The mass $\sf{m_1}$ goes downward and $\sf{m_2}$ goes upward when the system is released, since $\sf{m_1>m_2.}$

Then the net acceleration of the system is $\sf{a}$ in downward direction.

We know they move since they each have weight.

So the net force acting on the system will be,

$\sf{\longrightarrow F=m_1g-m_2g\quad\quad\dots(1)}$

Negative sign for $\sf{m_2g}$ is because the body is moving upward but its weight is downward.

By Second Law, the net force acting on the system is the product of net mass and net acceleration of the system.

$\sf{\longrightarrow F=(m_1+m_2)a\quad\quad\dots(2)}$

From (1) and (2),

$\sf{\longrightarrow (m_1-m_2)g=(m_1+m_2)a}$

$\sf{\longrightarrow a=\left(\dfrac{m_1-m_2}{m_1+m_2}\right)g}$

According to the question,

• $\sf{m_1=8\ kg}$

• $\sf{m_2=4\ kg}$

Then acceleration of each body will be,

$\sf{\longrightarrow a=\left(\dfrac{8-4}{8+4}\right)g}$

$\sf{\longrightarrow\underline{\underline{a=\dfrac{g}{3}\ m\,s^{-2}}}}$

Result:- The net acceleration of such system of masses $\sf{m}$ and $\sf{2m}$ is always $\sf{\dfrac{g}{3}\ m\,s^{-2}.}$