What kinetic energy is
Every moving object can do work before it comes to rest. The kinetic energy of a fast-flowing stream has been used to grind corn; sailing ships harness the kinetic energy of the wind. NCERT defines this directly: the kinetic energy of an object is a measure of the work the object can do by virtue of its motion. For a body of mass \(m\) moving with velocity \(\vec v\), the kinetic energy \(K\) is
$$ K \;=\; \tfrac12\, m\,(\vec v \cdot \vec v) \;=\; \tfrac12\, m v^2 \tag{5.5}$$
The quantity is "half the mass times the square of the speed". Because work and energy share the same dimensions, kinetic energy carries dimensions \([\text{ML}^2\text{T}^{-2}]\) and is measured in joules (J), named after James Prescott Joule. A 1 kg body moving at \(1~\text{m s}^{-1}\) has a kinetic energy of just \(0.5~\text{J}\); the \(v^2\) dependence means modest speeds already correspond to large energies.
A scalar, and never negative
NCERT states it plainly: "Kinetic energy is a scalar quantity." The cleanest way to see this is to write it as the dot product \(K=\tfrac12 m(\vec v\cdot\vec v)\). A dot product of a vector with itself returns a number, not a direction. Two cars of equal mass and equal speed travelling in opposite directions therefore carry exactly the same kinetic energy, even though their momenta are equal and opposite.
It follows that kinetic energy can never be negative. The mass \(m\) is positive and the speed-squared \(v^2\) is positive (or zero). The smallest value \(K\) can take is zero, reached only at the instant the body is momentarily at rest. The NIOS lesson poses exactly this as a check question — "Is it possible for a particle to have a negative value of kinetic energy?" — and the answer is no.
The work-energy theorem for a constant force
The theorem grows straight out of kinematics. NCERT §5.2 starts from the constant-acceleration relation for rectilinear motion,
$$ v^2 - u^2 = 2as \tag{5.2}$$
where \(u\) and \(v\) are the initial and final speeds and \(s\) the distance traversed. Multiply both sides by \(m/2\):
$$ \tfrac12 m v^2 - \tfrac12 m u^2 \;=\; m a s \;=\; F s $$
The last step uses Newton's second law, \(F=ma\). Generalising to three dimensions with vectors, \(v^2-u^2 = 2\,\vec a\cdot\vec d\), and again multiplying by \(m/2\) gives
$$ \tfrac12 m v^2 - \tfrac12 m u^2 \;=\; m\,\vec a\cdot\vec d \;=\; \vec F\cdot\vec d \tag{5.2b}$$
The left side is the change in the quantity "half the mass times the square of the speed" — the change in kinetic energy. The right side is the product of the displacement and the component of force along it — the work \(W\). Naming each side gives the theorem:
$$ \boxed{\,K_f - K_i \;=\; W_{\text{net}} \quad\Longleftrightarrow\quad \tfrac12 m v^2 - \tfrac12 m u^2 \;=\; W_{\text{net}}\,} \tag{5.3}$$
NIOS reaches the identical result by a parallel route, writing \(W=Fs=\left(m\dfrac{v_2^2-v_1^2}{2s}\right)s=\tfrac12 mv_2^2-\tfrac12 mv_1^2=K_2-K_1\), and states the theorem as: "the work done by the resultant of all forces acting on a body is equal to the change in kinetic energy of the body." The two textbooks agree — the word resultant (net) is essential.
Extension to a variable force
A constant force is rare; the variable force is more common. NCERT §5.6 proves the theorem still holds, confining the argument to one dimension. Start from the time-rate of change of kinetic energy:
$$ \frac{dK}{dt} \;=\; \frac{d}{dt}\!\left(\tfrac12 m v^2\right) \;=\; m v \frac{dv}{dt} \;=\; F v \;=\; F\frac{dx}{dt} $$
where Newton's second law supplies \(m\,dv/dt = F\). Cancelling \(dt\) gives \(dK = F\,dx\). Integrating from the initial position \(x_i\) to the final position \(x_f\):
$$ \int_{K_i}^{K_f} dK \;=\; \int_{x_i}^{x_f} F\,dx \qquad\Longrightarrow\qquad K_f - K_i \;=\; \int_{x_i}^{x_f} F\,dx \tag{5.8a}$$
The right-hand integral is exactly the definition of work done by a variable force — the area under the force-displacement curve. Hence
$$ K_f - K_i \;=\; W \tag{5.8b}$$
and the work-energy theorem is proved for a varying force. The form of the statement is unchanged; only the calculation of \(W\) becomes an integral instead of a product.
NCERT adds an important caveat that NEET likes to test: the theorem "does not, in general, incorporate the complete dynamical information of Newton's second law. It is an integral form of Newton's second law." Time information is integrated over and is not recoverable from \(W_{\text{net}}=\Delta K\); and because the result is a single scalar equation, the directional information present in the vector second law is absent.
The \(\int F\,dx\) area method is developed in full in work done by a variable force, including the spring and the trunk-pushing example.
Kinetic energy and momentum
Kinetic energy and linear momentum describe the same motion but are not interchangeable. For a body of mass \(m\) with momentum \(p=mv\), substitute \(v=p/m\) into \(K=\tfrac12 mv^2\):
$$ K \;=\; \tfrac12 m\left(\frac{p}{m}\right)^2 \;=\; \frac{p^2}{2m} \qquad\Longleftrightarrow\qquad p \;=\; \sqrt{2mK} $$
This relation is the engine behind NCERT Exercise 5.12, where an electron and a proton of given kinetic energies are compared. Reading the two faces of the relation carefully resolves most confusion:
| Quantity held fixed | Relation | Consequence |
|---|---|---|
| Same momentum \(p\) | \(K = \dfrac{p^2}{2m}\) | The lighter body has the larger kinetic energy |
| Same kinetic energy \(K\) | \(p = \sqrt{2mK}\) | The heavier body has the larger momentum |
| Same speed \(v\) | \(K=\tfrac12 mv^2,\; p=mv\) | The heavier body has both larger \(K\) and larger \(p\) |
How to read the work-energy theorem
The theorem is a bookkeeping statement about a single body. The net work done on it — the sum of works by every force acting, each positive, negative or zero — equals its change in kinetic energy. Three readings make it useful in NEET problems.
- Speeding up and slowing down. If the net work is positive the body gains kinetic energy and speeds up; if negative it loses kinetic energy and slows. A frictional or resistive force, opposing the displacement, does negative work and removes kinetic energy.
- Force need not be known in detail. Because only the net work matters, the theorem can be applied even when the exact force law is unknown. NCERT highlights this in the raindrop example, where the resistive force of air is undetermined, yet the work it does follows from the change in kinetic energy.
- It is the net force. The change in kinetic energy responds to the resultant of all forces. Adding the work of one force only — say, gravity — and ignoring friction will not give \(\Delta K\).
NCERT also notes that the theorem holds in all inertial frames, and can be extended to non-inertial frames provided pseudo-forces are included when computing the net force.
Worked example 1 — the raindrop (NCERT Example 5.2)
A raindrop of mass \(1.00~\text{g}\) falls from a height of \(1.00~\text{km}\). It hits the ground with a speed of \(50.0~\text{m s}^{-1}\). Taking \(g=10~\text{m s}^{-2}\), find (a) the work done by the gravitational force and (b) the work done by the unknown resistive force of air.
(a) Work done by gravity. Gravity acts over the full \(1000~\text{m}\) drop: \(W_g = mgh = (10^{-3})(10)(10^3) = 10.0~\text{J}\).
Change in kinetic energy. The drop starts from rest, so \(\Delta K = \tfrac12 m v^2 - 0 = \tfrac12 (10^{-3})(50)^2 = 1.25~\text{J}\).
(b) Work done by the resistive force. The work-energy theorem with the net work (gravity plus resistance) gives \(\Delta K = W_g + W_r\). Hence \(W_r = \Delta K - W_g = 1.25 - 10.0 = -8.75~\text{J}\).
Reading the result. \(W_r\) is negative — the air drains kinetic energy from the drop, which is why a raindrop arrives at a modest terminal-type speed rather than a lethal one. The exact force law is never needed; the theorem extracts the work from the energy balance alone.
Worked example 2 — electron vs proton (NCERT Exercise 5.12)
An electron and a proton are detected in a cosmic-ray experiment, the first with kinetic energy \(10~\text{keV}\) and the second with \(100~\text{keV}\). Which is faster, and what is the ratio of their speeds? Take electron mass \(m_e = 9.11\times10^{-31}~\text{kg}\), proton mass \(m_p = 1.67\times10^{-27}~\text{kg}\).
Set up from \(K=\tfrac12 mv^2\). Solving for speed, \(v=\sqrt{2K/m}\). The ratio of speeds is therefore
$$ \frac{v_e}{v_p} \;=\; \sqrt{\frac{2K_e/m_e}{2K_p/m_p}} \;=\; \sqrt{\frac{K_e}{K_p}\cdot\frac{m_p}{m_e}}. $$
Insert the numbers. The energy ratio is \(K_e/K_p = 10/100 = 0.1\). The mass ratio is \(m_p/m_e = (1.67\times10^{-27})/(9.11\times10^{-31}) \approx 1833\). So
$$ \frac{v_e}{v_p} \;=\; \sqrt{0.1 \times 1833} \;=\; \sqrt{183.3} \;\approx\; 13.5. $$
Conclusion. The electron is faster by a factor of about \(13.5\), even though it has only one-tenth the kinetic energy. Its far smaller mass dominates the comparison — a direct illustration that low kinetic energy does not imply low speed when masses differ greatly.
Worked example 3 — braking distance
A cyclist comes to a skidding stop in \(10~\text{m}\). The force on the cycle due to the road is \(200~\text{N}\), directly opposed to the motion. How much work does the road do on the cycle, and how does this connect to the work-energy theorem? (NCERT Example 5.3.)
Work done by the stopping force. The frictional force and the displacement are antiparallel, so \(\theta = 180^\circ\): \(W_r = Fd\cos\theta = 200 \times 10 \times \cos180^\circ = -2000~\text{J}\).
Reading via the theorem. This is the only horizontal force doing work, so \(\Delta K = W_r = -2000~\text{J}\). The cycle's kinetic energy falls by \(2000~\text{J}\) — exactly the amount needed to bring it from its initial speed to rest. The negative work brings the cycle to a halt, in accordance with the work-energy theorem.
Why braking distance grows as the square of speed. For a constant retarding force \(F\), the theorem gives \(Fd = \tfrac12 mv^2\), so the stopping distance \(d = \dfrac{mv^2}{2F} \propto v^2\). Doubling the entry speed quadruples the braking distance — a result that comes directly from the \(v^2\) in the kinetic-energy formula, and a recurring theme in road-safety physics.
Kinetic energy & the work-energy theorem in one breath
- \(K=\tfrac12 mv^2\) — energy of motion, a scalar, always \(\ge 0\), dimensions \([\text{ML}^2\text{T}^{-2}]\), unit joule.
- Work-energy theorem: \(W_{\text{net}} = K_f - K_i = \tfrac12 mv^2 - \tfrac12 mu^2\) — change in \(K\) equals work by the net force.
- Constant force: derived from \(v^2-u^2=2as\) times \(m/2\). Variable force: \(dK=F\,dx\) integrated gives \(K_f-K_i=\int F\,dx\).
- It is the integral, scalar form of Newton's second law — time and direction information are integrated over.
- Momentum link: \(K = p^2/2m\), or \(p=\sqrt{2mK}\). At fixed \(K\) the heavier body has more momentum; at fixed \(p\) the lighter body has more \(K\).
- Negative net work removes kinetic energy (friction, air resistance); braking distance \(\propto v^2\).