Answer
Verified
81.3k+ 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
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
A rope of 1 cm in diameter breaks if tension in it class 11 physics JEE_Main
Assertion The melting point Mn is more than that of class 11 chemistry JEE_Main