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Motion Of Connected Blocks Over Pulley MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Motion of connected blocks over pulley is considered one the most difficult concept.

  • 15 Questions around this concept.

Solve by difficulty

A light string passing over a smooth light pulley  connects two blocks of masses m1 and m2  ( vertically) .  If the acceleration of the system is g/8,  then the ratio of the masses is

Calculate the acceleration of the system as shown in the figure , assume all the surfaces are smooth 

 

When one block is hanging, the other is on the Table as shown in the below figure.

For this case what is the formula of tension force between them?

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Concepts Covered - 2

Motion of connected blocks over pulley

 

 

Equation of motion for   m_{1}

F_{net}=T-m_{1}g=m_{1}a

Equation of Motion for   m_{2}

F_{net}=m_{2}g-T=m_{2}a

a=\frac{[m_{2}-m_{1}\:]g}{m_{1}+m_{2}}

T=\frac{2m_{1}m_{2}\:g}{m_{1}+m_{2}}

When one Block is hanging from a rope and one on the table

 

  1. When one Block is hanging, other is on the Table

 

           

            a=\frac{m_{2}\:g}{m_{1}+m_{2}}


          T=\frac{m_{1}m_{2}\:g}{m_{1}+m_{2}}

 

  1. Three blocks, two are hanging and one is at the rest on the smooth horizontal table

   

 

      

m_1a=m_1g-T_1

m_2a=T_2- m_2\:g

T_1-T_2=Ma

 

a=\frac{(m_1-m_2)g}{m_1+m_2+M}


T_1=\frac{m_1(2m_2+M)\:g}{(m_1+m_2+M)}


T_2=\frac{m_2(2m_1+M)\:g}{(m_1+m_2+M)}

 

 

 

 

 

 

Study it with Videos

Motion of connected blocks over pulley
When one Block is hanging from a rope and one on the table

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