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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

QuestionEASY

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

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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 )                                                 

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

 

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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

Study other Related Concepts

Motion Of Connected Blocks Over Pulley Current Topic

Inertia
Forces
Common Forces In Mechanics
Equilibrium Of Concurrent Forces
Newton’s First Law Of Motion
Linear Momentum
Newton's Laws Of Motion
Acceleration Of Block On Horizontal Smooth Surface
Acceleration Of Block On Smooth Inclined Plane
Motion Of Bodies In Contact

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