
Prove work energy theorem for a constant force.
Answer
134.1k+ views
Hint: Work energy theorem gives the relation between work done and energy. According to the work energy theorem, the net work done on a body is equal to the change in the kinetic energy of the body.
Complete step by step solution:
Suppose an object is having a mass ‘m’. Initially the object is moving with a velocity \[{v_1}\] and its final velocity is \[{v_2}\].
Therefore the initial kinetic energy of the object will be \[{K_1} = \dfrac{1}{2}mv_1^2\].
The final kinetic energy of the object will be \[{K_2} = \dfrac{1}{2}mv_2^2\].
Given that a constant force is acting on the object, so using Newton’s second law of motion, it can be written that
F=m.a……(i)
Where ‘F’ is the force, ‘m’ is the mass and ‘a’ is the acceleration
Also work done is defined as the product of force applied and the displacement. Mathematically, work done is written as
W=F.d……(ii)
It is known that the acceleration is the rate of change of velocity of the object. If the velocity of the object is changing and the object covers a displacement ‘d’, then using the equation of motion we can write that
\[v_2^2 - v_1^2 = 2ad\]
\[\Rightarrow a = \dfrac{{v_2^2 - v_1^2}}{{2d}}\]
Substituting the value of acceleration in equation (i) and solving, we get
\[F = m.\dfrac{{v_2^2 - v_1^2}}{{2d}}\]
\[\Rightarrow F.d = \dfrac{1}{2}m(v_2^2 - v_1^2)\]
Using equation (ii), in the above equation we get
\[W = \dfrac{1}{2}m(v_2^2 - v_1^2)\]
\[\Rightarrow W = \Delta K.E.\]
Where ‘W’ is the work done and \[\Delta K.E.\] is the kinetic energy.
Hence Proved
Note: It is important to remember that work energy is used to find out the work done by a number of forces on a solid object if it is moving under the influence of a number of forces. Work energy theorem is scalar as it does not define the direction of velocity in which the object is moving.
Complete step by step solution:
Suppose an object is having a mass ‘m’. Initially the object is moving with a velocity \[{v_1}\] and its final velocity is \[{v_2}\].
Therefore the initial kinetic energy of the object will be \[{K_1} = \dfrac{1}{2}mv_1^2\].
The final kinetic energy of the object will be \[{K_2} = \dfrac{1}{2}mv_2^2\].
Given that a constant force is acting on the object, so using Newton’s second law of motion, it can be written that
F=m.a……(i)
Where ‘F’ is the force, ‘m’ is the mass and ‘a’ is the acceleration
Also work done is defined as the product of force applied and the displacement. Mathematically, work done is written as
W=F.d……(ii)
It is known that the acceleration is the rate of change of velocity of the object. If the velocity of the object is changing and the object covers a displacement ‘d’, then using the equation of motion we can write that
\[v_2^2 - v_1^2 = 2ad\]
\[\Rightarrow a = \dfrac{{v_2^2 - v_1^2}}{{2d}}\]
Substituting the value of acceleration in equation (i) and solving, we get
\[F = m.\dfrac{{v_2^2 - v_1^2}}{{2d}}\]
\[\Rightarrow F.d = \dfrac{1}{2}m(v_2^2 - v_1^2)\]
Using equation (ii), in the above equation we get
\[W = \dfrac{1}{2}m(v_2^2 - v_1^2)\]
\[\Rightarrow W = \Delta K.E.\]
Where ‘W’ is the work done and \[\Delta K.E.\] is the kinetic energy.
Hence Proved
Note: It is important to remember that work energy is used to find out the work done by a number of forces on a solid object if it is moving under the influence of a number of forces. Work energy theorem is scalar as it does not define the direction of velocity in which the object is moving.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3
