
An object of mass 16 $kg$ is moving with an acceleration of 3 $m/{s^2}$. Calculate the applied force. If the same force is applied on an object of mass 24 $kg$, how much will be the acceleration?
Answer
217.2k+ views
Hint: In the first case, the value of the mass of an object and value of the acceleration of an object is given and applied force is asked. Use Newton’s second law of motion i.e., ${\text{F = ma}}$ and then substitute the given values in the formula and evaluate the value of the applied force. Now in the second case, the value of applied force in both the cases are the same. Value of mass of an object is given. Again using Newton’s second law of motion evaluate the value of acceleration of an object.
Complete step by step solution:
Given: Mass of an object, ${{\text{m}}_{\text{1}}}{\text{ = 16 kg}}$
Acceleration of the object, ${{\text{a}}_{\text{1}}}{\text{ = 3 m/}}{{\text{s}}^{\text{2}}}$
To find: Applied force
Using Newton’s second law of motion
${\text{F = ma}}$
Where F = Applied force
M = Mass of the object
And a = Acceleration of the object
Now substituting the given values in above formula, we get
${{\text{F}}_{\text{1}}}{\text{ = }}{{\text{m}}_{\text{1}}}{{\text{a}}_{\text{1}}}{\text{ = 16 \times 3 = 48 N}}$
Given that applied force on first object is same as that of the applied force on the another object
$\therefore {\text{ }}{{\text{F}}_{\text{1}}}{\text{ = }}{{\text{F}}_{\text{2}}}{\text{ = 48 N}}$
Mass of another object, ${{\text{m}}_{\text{2}}}{\text{ = 24 kg}}$
Acceleration of the object =?
Again using Newton’s second law of motion
${\text{F = ma}}$
Where F = Applied force
M = Mass of the object
And a = Acceleration of the object
Now substituting the given values in above formula, we get
$
{F_2}{\text{ = }}{{\text{m}}_{\text{2}}}{{\text{a}}_{\text{2}}} \\
\Rightarrow {{\text{a}}_{\text{2}}}{\text{ = }}\dfrac{{{{\text{F}}_{\text{2}}}}}{{{{\text{m}}_{\text{2}}}}}{\text{ }} = {\text{ }}\dfrac{{48}}{{24}}{\text{ = 3 m/}}{{\text{s}}^2} \\
$
Therefore, applied force on an object is 48 N and acceleration of another object is 3 $m/{s^2}$.
Note: The second law states that the rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force.
\[{\text{F = }}\dfrac{{{\text{dp}}}}{{{\text{dt}}}}{\text{ = }}\dfrac{{{\text{d(mv)}}}}{{{\text{dt}}}}{\text{ = m}}\dfrac{{{\text{dv}}}}{{{\text{dt}}}}{\text{ = ma}}\]
Complete step by step solution:
Given: Mass of an object, ${{\text{m}}_{\text{1}}}{\text{ = 16 kg}}$
Acceleration of the object, ${{\text{a}}_{\text{1}}}{\text{ = 3 m/}}{{\text{s}}^{\text{2}}}$
To find: Applied force
Using Newton’s second law of motion
${\text{F = ma}}$
Where F = Applied force
M = Mass of the object
And a = Acceleration of the object
Now substituting the given values in above formula, we get
${{\text{F}}_{\text{1}}}{\text{ = }}{{\text{m}}_{\text{1}}}{{\text{a}}_{\text{1}}}{\text{ = 16 \times 3 = 48 N}}$
Given that applied force on first object is same as that of the applied force on the another object
$\therefore {\text{ }}{{\text{F}}_{\text{1}}}{\text{ = }}{{\text{F}}_{\text{2}}}{\text{ = 48 N}}$
Mass of another object, ${{\text{m}}_{\text{2}}}{\text{ = 24 kg}}$
Acceleration of the object =?
Again using Newton’s second law of motion
${\text{F = ma}}$
Where F = Applied force
M = Mass of the object
And a = Acceleration of the object
Now substituting the given values in above formula, we get
$
{F_2}{\text{ = }}{{\text{m}}_{\text{2}}}{{\text{a}}_{\text{2}}} \\
\Rightarrow {{\text{a}}_{\text{2}}}{\text{ = }}\dfrac{{{{\text{F}}_{\text{2}}}}}{{{{\text{m}}_{\text{2}}}}}{\text{ }} = {\text{ }}\dfrac{{48}}{{24}}{\text{ = 3 m/}}{{\text{s}}^2} \\
$
Therefore, applied force on an object is 48 N and acceleration of another object is 3 $m/{s^2}$.
Note: The second law states that the rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force.
\[{\text{F = }}\dfrac{{{\text{dp}}}}{{{\text{dt}}}}{\text{ = }}\dfrac{{{\text{d(mv)}}}}{{{\text{dt}}}}{\text{ = m}}\dfrac{{{\text{dv}}}}{{{\text{dt}}}}{\text{ = ma}}\]
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

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

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Understanding Atomic Structure for Beginners

