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Motion Of Connected Blocks Over Pulley - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

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

  • 18 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

Two masses of 10 kg and 9 kg are connected by a string that passes over a smooth pulley; What is the tension in the string \left ( g=9.81ms^{-2} \right )                                                 

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

 

Concepts Covered - 2

Motion of connected blocks over pulley

 

 

Equation of motion for $m_1$

$
F_{n e t}=T-m_1 g=m_1 a
$


Equation of Motion for $m_2$

$
\begin{aligned}
& F_{\text {net }}=m_2 g-T=m_2 a \\
& a=\frac{\left[m_2-m_1\right] g}{m_1+m_2} \\
& T=\frac{2 m_1 m_2 g}{m_1+m_2}
\end{aligned}
$
 

When one Block is hanging from a rope and one on the table
  1. When one Block is hanging, other is on the Table

 

           

     $\begin{aligned} a & =\frac{m_2 g}{m_1+m_2} \\ T & =\frac{m_1 m_2 g}{m_1+m_2}\end{aligned}$

 

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

   

 

      

$
\begin{aligned}
& m_1 a=m_1 g-T_1 \\
& m_2 a=T_2-m_2 g \\
& T_1-T_2=M a
\end{aligned}
$


$
\begin{aligned}
& a=\frac{\left(m_1-m_2\right) g}{m_1+m_2+M} \\
& T_1=\frac{m_1\left(2 m_2+M\right) g}{\left(m_1+m_2+M\right)} \\
& T_2=\frac{m_2\left(2 m_1+M\right) g}{\left(m_1+m_2+M\right)}
\end{aligned}
$
 

 

 

 

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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