6.3.4 WORK-ENERGY THEOREM
State and prove work-energy theorem for a constant force.
Work-Energy Theorem : This theorem states that the work done by a force to move a body is equal to the change in kinetic energy of the body.
work done by a force on body = change in kinetic energy of the body$
Proof :
Consider a body of mass $m$ moving with a velocity $u$. Let a force $F$ be applied on the body, so that it is accelerated with an acceleration '$a$'.
Then,
$$F = ma$$If $S$ be the distance travelled by the body during its accelerated motion in the direction of applied force, then the work done by the force $F$ on the body is given by
$$W = FS = ma \, S \quad \quad \quad (\because F = ma) \quad \dots(1)$$Let the body acquires velocity $v$ after travelling a distance $S$, then from $v^2 - u^2 = 2 \, aS$, we have
$$a = \frac{v^2 - u^2}{2 \, S} \quad \dots(2)$$Put this value in eqn. (1), we get
$$W = m \left( \frac{v^2 - u^2}{2 \, S} \right) \times S = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$ $$W= K_f - K_i \quad \dots(3)$$where, $\frac{1}{2}m \, v^2 = \text{Final K.E., } (K_f)$ and $\frac{1}{2}m \, u^2 = \text{Initial K.E., } (K_i)$
The difference between the final and initial kinetic energies is the change in K.E. of the body ($\Delta K$), where $\Delta$ means change.
$\therefore W = \text{Change in K.E. of a body} = \Delta K$
This is known as work-energy theorem.
Discussion of Work-Energy Theorem :
(i) Work done by a force is zero, if there is no change in the speed of a particle or a body.
Explanation : $W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$
If $u = v$, we get
$$W = 0$$Example : When a particle moves in a circular path with constant speed, there is no change in the kinetic energy of the particle and hence according to work-energy theorem, work done on the particle is zero.
(ii) Work done by a force is negative, if there is decrease in the speed of a particle or a body.
Explanation : $W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2}m(v^2 - u^2)$
Since v u
$$\therefore W = \text{Negative.}$$Example : When a particle is projected upward, then the work done by the gravitational force is negative. In this case, as the particle moves up, its speed decreases and hence consequently the kinetic energy of the particle also decreases.
(iii) Work done by the force is positive, if there is increase in the velocity of the particle.
Explanation : $W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2}m(v^2 - u^2)$
Since $v > u$
$$\therefore W = \text{Positive.}$$Example : When a particle or an object is dropped from the top of a building, then the work done by the gravitational force is positive. As the particle descends, its speed increases and hence kinetic energy of the particle also increases. This implies that the work done by the gravitational force is positive.
Frequently Asked Questions (FAQ)
Q1: What happens to the kinetic energy if the work done by the net force is negative?
Answer: If the work done is negative, it means the force is acting opposite to the direction of motion (like friction). According to the theorem, negative work results in a decrease in the body's kinetic energy, causing it to slow down.
Q2: Does the Work-Energy Theorem apply to variable forces?
Answer: Yes, while the proof above uses a constant force, the Work-Energy Theorem holds true for variable forces as well. For variable forces, calculus integration ($\int F \, dx$) is used to find total work done, which still equals the total change in kinetic energy.
Q3: Is the Work-Energy Theorem applicable in the presence of friction?
Answer: Yes, it is. The term "work done" refers to the work done by all forces acting on the body, including conservative forces (like gravity) and non-conservative forces (like friction or air resistance).
Quick Quiz (5 MCQs)
Test your understanding of the core definitions, proofs, and cases discussed above:
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