
Prove work energy theorem for a constant force.
Answer
233.1k+ 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 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

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

