Suppose a body is moving on a smooth horizontal surface with a constant velocity, u. let a constant force F acts on the body from point A and B as shown in the figure, such that the velocity increases to v. the work done by the force is
W = F\(\times\) s where s is the displacement of the body.
From Newton’s second law of motion,
F = m\(\times\)a where m is the mass of the body.
Then,,
W = mas
From the equation of motion
v^{2} = u^{2} + 2as,
v^{2} –u^{2} = 2as
or, as = \(\frac 12( {v^2 –u^2}\))
Work done, = ma s =m \(\frac {1}{2}\)(v^{2}-u^{2}) = \(\frac {1}{2}\)m.v^{2} - \(\frac {1}{2}\) mu^{2}
or, W=(K.E.)f – (K.E.)i
So, the work done on moving the body by the force is equal to the increase in its kinetic energy. Again we can write the above equation as
W = \(\frac {1}{2}\) m.v^{2} - \(\frac {1}{2}\) m.u^{2}
or, \(\frac {1}{2}\) mv^{2} = W + \(\frac {1}{2}\) m.u^{2}
or,(K.E.)f = W + (K.E.)i
That is, the final kinetic energy of the body is increased and it is the sum of the work done on the body by the applied force and its initial kinetic energy. This principle is called work-energy theorem. If the force acts in opposite to the direction of the motion of the body, the force does the negative work and it is given as
\begin{align*} -W &= \frac 12 mv^2 - \frac 12 mu^2 \\ \text {or,} \frac 12 mv^2 &= \frac 12 mu^2 – W \\ \end{align*}
So, the final kinetic energy decreases.
According to this principle, energy of an isolated system is constant. It is neither be created nor be destroyed but it transforms from one type to another type. When an object falls on the ground from the certain height, its kinetic energy is converted into another form of energy such as sound energy, heat energy, light energy etc.
Energy Conservation for a Freely Falling Body
The mechanical energy of a freely falling body under the gravity is constant. This is shown in the figure.
Let us take an object of mass m at a point A, above the ground at a height of h from the ground.
At A, the body is at rest and
\begin{align*} \text {K.E. of the body} &= 0 \\ \text {P.E. of the body} &= mgh \\ \therefore \text {Total energy of the body at A} &= KE + PE = 0 + mgh = mgh \\ \end{align*}
Let the body falls freely from A to the ground and at a point B on its path, its height above the ground will become (h-x).
\begin{align*} \therefore \text {potential energy of the body} &= mg(h-x) \\ \text {if} V_b \text {is the velocity at B, then} \\ V_B^2 &= u^2 + 2as = 0 + 2gx \\ \text {or,} V_B^2 &= 2gx \\ \therefore \text {K.E. of the body} &= \frac 12 mV_b^2 = \frac 12 m \times 2gx = m\: gx \\ \text {At point C} \\ \end{align*} If V_{C} be the velocity of the body at point C just before striking the ground .
\begin{align*} \text {then} \\ V_C^2 &= 0 + 2gh \\ V_C^2 &= 2gh \\ \text {and} \: \text {K.E. of the body} &= \frac 12 mV_C^2 = \frac 12 m \times 2gh = m\: gh \\ \text {and} \: \text {P.E.} &= m\: gh = 0 \\ \therefore \text {Total energy at point C} = KE + PE = m\: gh + 0 = m\: gh \\ \end{align*}
Thus, the total mechanical energy of the body remains the same at all points during the downward journey. As the body falls from the height h, its potential energy decreases while kinetic energy begins to increases; the total mechanical energy at any point being the same i.e. m gh. When the body reaches the ground, P.E. = 0 and K.E. is equal to the initial potential energy m gh at the height h. figure shows the variation of P.E. and K.E. with height. Similar case can be applied when a body is thrown vertically upward.
A force is said to be conservative if the work done by or against the force of moving a body depends only on the initial and final positions of the body, and not on the nature of path followed between the initial and the final positions i.e. amount of work done is independent of path followed. It means work done by or against a conservative force in moving a body any path between fixed and final positions will be the same.
Gravitational force, force in an elastic spring, electrostatic force, magnetic force etc. are examples of conservative forces.
In the case of gravitational force, if we take work done in moving the body from Y to X, by gravity has to be taken as positive. So, they are equal and opposite i.e.
\begin{align*} W_{XY} &= -W_{YX} \text {or,} \: W_{XY} + W_{YX} = 0 \end{align*}
It means, the total amount of work done in a closed path in the conservative force field (gravitational field) is zero.
Properties of Conservative Forces
Non-Conservative Forces
A force is said to be non-conservative, if work done by or against the force in moving a body from one position to another, depends on the path followed between these two positions.
Frictional forces are non-conservative forces. If a body is moved from position A to B on a rough table, work done against frictional force shall depend on the length of the path between A and B and not only on the positions A and B.
Power is defined as the rate at which the work is done.
$$\text {Power} = \frac {\text {work}}{\text {time}} = \frac Wt $$
Thus power of an agent measures how fast it can do the work.
For constant force,
$$\text {Power, P} = \frac Wt = \frac {\vec F . \vec s}{t} = \vec F.\vec v $$
Where \(\vec v = \frac {\vec s}{t} \), linear velocity.
Hence the power of an agent can be expressed as the dot product of the force applied and velocity of the body.
If Ï´ is the angle between \(\vec F \text {and} \vec v\), then
$$ P = \vec F. \vec v = Fv\cos \theta $$
However, when \(\vec v\) is along \(\vec F\), Ï´ = 0^o and
$$ P = Fv\cos 0^o = Fv$$
ASK ANY QUESTION ON Work-Energy Theorem, Principle of Conservation of Energy and Types of Forces
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sujan
work energy theorem
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