Question

Two blocks of masses 6 kg and 4 kg are attached to the two ends of a massless string passing over a smooth fixed pulley. If the system is released, the acceleration of the centre of mass of the system will be:
(A) g, vertical downwards
(B) \[\dfrac{g}{5}\], vertical downwards
(C) \[\dfrac{g}{{25}}\], vertical downwards
(D) zero

Answer 1

Hint:Here,we are going to apply the concept of the Atwood machine and after putting the values of masses,we can arrive at the correct solution.

Complete step by step answer:
We know that the basic formula for Atwood machine is,
\[\Rightarrow a = \dfrac{{\left( {{m_1} - {m_2}} \right)}}{{{m_1} + {m_2}}}g\]
Where \[{m_1}\]= 6kg
\[\Rightarrow {m_2}\]= 4kg
g is the gravitational force
Now putting the value in the above expression, we get,
\[\Rightarrow a = \dfrac{{\left( {6 - 4} \right)}}{{6 + 4}}g\]
Further simplifying it we get,
\[\Rightarrow a = \dfrac{{\left( 2 \right)}}{{10}}g\]
Now by dividing we get,
\[\Rightarrow a = 0.2g\]
Now the acceleration of the centre of mass will be,
\[\Rightarrow a = \dfrac{{\left( {{m_1}{a_1} - {m_2}{a_2}} \right)}}{{{m_1} + {m_2}}}g\]
Putting the values, we get,
\[\Rightarrow a = \dfrac{{ - \left( {6 - 4} \right)0.2}}{{6 + 4}}g\]
6kg mass is going down and 4kg mass is going up, hence acceleration of 4kg can be taken as negative.
Now by further simplifying we get,
\[\Rightarrow a = \dfrac{{ - \left( 2 \right)0.2}}{{10}}g\]
\[\Rightarrow a = 0.04g\]by further simplification
We can also write the above answer as,
\[\Rightarrow
a = - \dfrac{g}{{25}}\]
Therefore, the correct option is c).
Additional information:
A free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of \[9.8m/{s^2}\]downward. The numerical value for the acceleration of a free-falling object is a particularly important value that is given a specific name. It is known or referred to as the acceleration due to gravity.


Note: Students must remember to take acceleration of 4kg can be taken as negative otherwise the whole calculation will be wrong. In many cases acceleration can be negative when any body moves upwards against the force of acceleration which acts downwards.