
A constant force $F ={m_1}g/2$ is applied on the block of mass ${m_2}$ as shown in the figure. The string and pulley are light and the surface of the table is smooth. Find the acceleration of ${m_2}$.

Answer
216k+ views
Hint: Newton’s second law will be applied in this problem.
F = ma, force is equal to mass and acceleration; net force applied on an object because there are other forces also acting on an object such as frictional force, tension and gravitational force.
If net force is greater than other acting forces on the body acceleration in the body will take place.
Using the above mentioned concept we will calculate the acceleration of mass ${m_2}$.
Complete step by step solution:
Newton’s Second Law: It states that force is equal to the change in momentum and change in time.
$F = \dfrac{{dP}}{{dt}}$ (P is the momentum of the body)
Where P = m v (m is the mass of the body and v is the velocity of the body)
When change in velocity takes place acceleration is developed, therefore our equation becomes:
$F = \dfrac{{mdv}}{{dt}}$
$F = ma$
Let’s conclude the calculation part of the question now.
Force given to us, F = ${m_1}$g/2
Our equation becomes;
$ \Rightarrow T - F = {m_2}a$ (T is the tension of the string; F is the force we have applied)
$
\Rightarrow {m_2}a = T - F \\
\Rightarrow {m_2}a = T - \dfrac{{{m_1}g}}{2} \\
$ (We have substituted the given value)....................(1)
Forces acting on mass ${m_1}$ are;
${m_1}a = {m_{1g}} - T$ ...........................(2)
On adding equation (1) and (2)
$
\Rightarrow {m_1}a + {m_2}a = T - \dfrac{{{m_1}g}}{2} + {m_1}g - T \\
\Rightarrow a({m_1} + {m_2}) = - \dfrac{{{m_1}g}}{2} + {m_1}g \\
$
(Cancelling the T on RHS and taking a common from LHS)
$\Rightarrow a({m_1} + {m_2}) = \dfrac{{ - {m_1}g + 2{m_1}g}}{2} \\
\Rightarrow a = \dfrac{{{m_1}g}}{{2\left( {{m_1} + {m_2}} \right)}} \\
$ (Acceleration of mass ${m_2}$)
Note: We have a number of daily life examples where we are observing Newton’s second law of motion such as, hitting the golf ball, cricket ball, football with a force the more we exert force the more we will observe the reaction. As in our question above force having higher magnitude was existing in the system.
F = ma, force is equal to mass and acceleration; net force applied on an object because there are other forces also acting on an object such as frictional force, tension and gravitational force.
If net force is greater than other acting forces on the body acceleration in the body will take place.
Using the above mentioned concept we will calculate the acceleration of mass ${m_2}$.
Complete step by step solution:
Newton’s Second Law: It states that force is equal to the change in momentum and change in time.
$F = \dfrac{{dP}}{{dt}}$ (P is the momentum of the body)
Where P = m v (m is the mass of the body and v is the velocity of the body)
When change in velocity takes place acceleration is developed, therefore our equation becomes:
$F = \dfrac{{mdv}}{{dt}}$
$F = ma$
Let’s conclude the calculation part of the question now.
Force given to us, F = ${m_1}$g/2
Our equation becomes;
$ \Rightarrow T - F = {m_2}a$ (T is the tension of the string; F is the force we have applied)
$
\Rightarrow {m_2}a = T - F \\
\Rightarrow {m_2}a = T - \dfrac{{{m_1}g}}{2} \\
$ (We have substituted the given value)....................(1)
Forces acting on mass ${m_1}$ are;
${m_1}a = {m_{1g}} - T$ ...........................(2)
On adding equation (1) and (2)
$
\Rightarrow {m_1}a + {m_2}a = T - \dfrac{{{m_1}g}}{2} + {m_1}g - T \\
\Rightarrow a({m_1} + {m_2}) = - \dfrac{{{m_1}g}}{2} + {m_1}g \\
$
(Cancelling the T on RHS and taking a common from LHS)
$\Rightarrow a({m_1} + {m_2}) = \dfrac{{ - {m_1}g + 2{m_1}g}}{2} \\
\Rightarrow a = \dfrac{{{m_1}g}}{{2\left( {{m_1} + {m_2}} \right)}} \\
$ (Acceleration of mass ${m_2}$)
Note: We have a number of daily life examples where we are observing Newton’s second law of motion such as, hitting the golf ball, cricket ball, football with a force the more we exert force the more we will observe the reaction. As in our question above force having higher magnitude was existing in the system.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

