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Mechanics
Kinetic Energy in Classical Mechanics
Kinetic energy is the scalar energy possessed by a body due to its translational or rotational motion. It represents the total work required to accelerate a body from rest to its current velocity.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition, units, dimensional formula, work-energy theorem for rectilinear motion, simple numericals involving mass and velocity, and relation between linear momentum and kinetic energy.
PGT extension
Relativistic kinetic energy, rotational kinetic energy, rolling motion combining translation and rotation, work-energy theorem in non-inertial frames, and kinetic energy expressed in generalized coordinates.
Prerequisites
Newton's laws of motion
Work and scalar product of vectors
Conservative and non-conservative forces
Definition and Physical Nature
Kinetic energy (KE) is a scalar quantity dependent on the reference frame.
Its dimensional formula is [M L2 T-2] and the SI unit is the Joule (J).
It is always non-negative for a real particle in classical mechanics.
Work-Energy Theorem
The net work done by all forces acting on a particle equals the change in its kinetic energy.
Expressed mathematically as Wnet = ΔK$E = KE_{final} - KE_{initial}$.
Holds true for both constant and variable forces in inertial reference frames.
Relation with Linear Momentum
For a particle of mass m moving with linear momentum p, K$E = p^{2} / (2m)$.
If momentum is doubled at constant mass, kinetic energy increases by a factor of four.
For a system of particles, total kinetic energy is the sum of translational KE of the centre of mass and internal KE relative to the centre of mass.
Rotational and Combined Kinetic Energy
A rigid body rotating about a fixed axis possesses rotational kinetic energy equal to (1/2)Iω2.
A body undergoing simultaneous translation and rotation has total kinetic energy K$E = (1/2)Mv_{cm}^{2} + (1/2)I_{cm}\omega^{2}$.
In relativistic mechanics, kinetic energy is given by K$E = (\gamma - 1)mc^{2}$.
Derivations
Work-Energy Relation for Rectilinear Motion
Conditions: Particle of mass m moving along a straight line under a net force F
Start with Newton's second law: $F = m(dv/dt) = m(dv/dx)(dx/dt) = mv(dv/dx)$.
Integrate force with respect displacement x from initial position x1 to final position x2: $W = \int F dx = \int mv (dv/dx) dx$.
Change the integration variable from displacement to velocity: $W = \int _{v1}^{v_{2}} mv dv$.
Evaluate the integral: $W = m [(v^{2}/2)]_{v1}^{v_{2}} = (1/2)mv_{2}^{2} - (1/2)mv_{1}^{2}$.
Kinetic Energy versus Velocity (Classical vs Relativistic)
Draw: A 2D plot with velocity v on the horizontal axis and kinetic energy on the vertical axis. Draw a parabolic curve for classical K$E = (1/2)mv^2$ and an asymptotically rising curve for relativistic KE.
Interpret: Classical KE grows quadratically with velocity and is valid for v << c. Relativistic KE approaches infinity as v approaches c.
Exam use: Tested in conceptual multiple-choice questions regarding validity limits of classical mechanics.
Kinetic Energy versus Momentum
Draw: A parabolic curve plotting KE on the vertical axis against momentum p on the horizontal axis for a constant mass.
Interpret: Since K$E = p^2 / (2m)$, the curve is a parabola opening upwards with slope dependent on 1/(2m).
Exam use: Frequently used in numerical problems evaluating percentage change in kinetic energy when momentum is altered.
Draw: Schematic diagram of a trolley rolling down an inclined track with photogates measuring velocity at two points.
Interpret: Verifies that the work done by the component of gravitational force along the incline equals the experimentally measured gain in translational kinetic energy.
A solid cylinder of mass 4 kg and radius 0.1 m rolls without slipping on a horizontal surface with a translational velocity of 2 m/s. Calculate its total kinetic energy.
Identify given parameters: M = 4 kg, R = 0.1 m, $v = 2 m/s$.
For a solid cylinder, the moment of inertia about its central axis is $I = (1/2) M R^2$.
For rolling without slipping, angular velocity $\omega = v / R$.
Total kinetic energy is K$E = (1/2) M v^2 + (1/2) I \omega^2 = (1/2) M v^2 + (1/2) [(1/2) M R^2] (v / R)^2 = (1/2) M v^2 + (1/4) M v^2 = (3/4) M v^2$.
Classical kinetic energy formula ((1/2)mv^2) assumes speeds are significantly smaller than the speed of light in vacuum (v << c).
Assumes the observer is in an inertial frame of reference; in non-inertial frames, pseudo forces must be accounted for in the work-energy theorem.
Common misconceptions
Common error: Kinetic energy is an absolute property of a body regardless of the observer.
Correct: Kinetic energy is frame-dependent and varies depending on the velocity of the observer's reference frame.
Velocity depends on the chosen frame of reference, hence K$E = (1/2)mv^2 yields different values$ for different observers.
Common error: If the net work done on a body is zero, its velocity must be zero.
Correct: If net work is zero, the change in kinetic energy is zero, meaning speed remains constant, but velocity vector may change direction (as in uniform circular motion).
Work depends on displacement and force along the displacement, whereas speed can remain constant while direction changes.
Quick revision
Kinetic energy is a scalar quantity given by K$E = (1/2)mv^2 = p^2/(2m) in classical mechanics$.
Translational Kinetic Energy: K$E = \frac{1}{2} m v^2$
Kinetic Energy from Momentum: K$E = \frac{p^2}{2m}$
Must remember
Kinetic energy (KE) is a scalar quantity and its dimensional formula is $[M L^2 T^{-2}]$ with SI unit Joule (J).
The work-energy theorem states that net work done equals the change in kinetic energy ($W_{net} = \Delta KE$).
Relation between linear momentum ($p$) and kinetic energy for a particle of mass $m$ is $KE = \frac{p^2}{2m}$.
For a rigid body undergoing simultaneous translation and rolling motion, $KE = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} I_{cm} \omega^2$.
Kinetic energy is frame-dependent and always non-negative for a real particle in classical mechanics.
Glossary
Kinetic EnergyThe energy possessed by a body by virtue of its motion, given by $\frac{1}{2}mv^2$ in classical mechanics.
Work-Energy TheoremA fundamental principle stating that the net work done on an object equals its change in kinetic energy.
Rotational Kinetic EnergyEnergy due to rotation about an axis, expressed as $\frac{1}{2}I\omega^2$.
Relativistic Kinetic EnergyExpression for kinetic energy at speeds approaching the speed of light, given by $(\gamma - 1)mc^2$.
Classical vs Relativistic Kinetic Energy
Parameter
Classical Mechanics
Relativistic Mechanics
Formula
$KE = \frac{1}{2}mv^2$
$KE = (\gamma - 1)mc^2$
Speed Limit
Valid for $v \ll c$
Valid for speeds comparable to $c$
Mass dependence
Mass is constant
Accounts for relativistic factor $\gamma$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the dimensional formula of kinetic energy?
Explanation: Kinetic energy has the same dimensional formula as work and energy, which is $[M L^2 T^{-2}]$.
2. If the linear momentum of a body is doubled while its mass remains constant, its kinetic energy will increase by a factor of:
Explanation: Since $KE = \frac{p^2}{2m}$, if $p$ becomes $2p$, $KE$ becomes $\frac{(2p)^2}{2m} = 4 \left(\frac{p^2}{2m}\right)$, meaning it increases by a factor of 4.
3. Which of the following quantities is frame-dependent in classical mechanics?
Explanation: Kinetic energy depends on the velocity of the object relative to the reference frame, making it frame-dependent.
4. The work-energy theorem is mathematically expressed as:
Explanation: The work-energy theorem states that the net work done by all forces equals the change in kinetic energy: $W_{net} = \Delta KE$.
5. What is the rotational kinetic energy of a rigid body rotating with angular velocity $\omega$ and moment of inertia $I$?
Explanation: Rotational kinetic energy is given by the standard expression $KE_{rot} = \frac{1}{2}I\omega^2$.
6. For a rolling body without slipping, the total kinetic energy is the sum of:
Explanation: Total kinetic energy of rolling motion is $KE = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$.
7. What is the correct relativistic expression for kinetic energy?
Explanation: In relativistic mechanics, total energy is $\gamma mc^2$ and rest energy is $mc^2$, so kinetic energy is $KE = (\gamma - 1)mc^2$.
8. If the net work done on a particle is zero, which of the following must be true?
Explanation: By the work-energy theorem, if $W_{net} = 0$, then $\Delta KE = 0$, meaning speed remains constant.
9. What is the SI unit of kinetic energy?
Explanation: Kinetic energy is a form of energy, and its SI unit is the Joule (J).
10. Instantaneous power delivered to a particle is related to its kinetic energy by:
Explanation: Power is the rate of doing work, which equals the rate of change of kinetic energy: $P = \frac{d(KE)}{dt} = \vec{F} \cdot \vec{v}$.
11. For a system of particles, the total kinetic energy can be expressed as the translational kinetic energy of the centre of mass plus:
Explanation: Total KE for a system of particles is $KE = \frac{1}{2}Mv_{cm}^2 + \sum \frac{1}{2}m_i v_i'^2$ (internal kinetic energy relative to CM).
12. Can the kinetic energy of a real particle ever be negative in classical mechanics?
Explanation: Since mass $m$ is positive and $v^2$ is always positive or zero, kinetic energy $\frac{1}{2}mv^2$ is always non-negative.
13. If momentum of a particle is $p$ and mass is $m$, kinetic energy is expressed as:
2. If the kinetic energy of a particle is increased by $300%$, the percentage increase in its linear momentum is:
Explanation: If $KE$ increases by $300%$, final $KE_2 = KE_1 + 3KE_1 = 4KE_1$. Since $p \propto \sqrt{KE}$, $p_2 = \sqrt{4}p_1 = 2p_1$. Percentage increase in momentum is $\frac{2p_1 - p_1}{p_1} \times 100 = 100%$.
3. A solid sphere of mass $M$ and radius $R$ rolls without slipping on a horizontal surface with linear velocity $v$. Its total kinetic energy is:
Explanation: For a solid sphere, $I_{cm} = \frac{2}{5}MR^2$ and $\omega = \frac{v}{R}$. $KE = \frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{v}{R}\right)^2 = \left(\frac{1}{2} + \frac{1}{5}\right)Mv^2 = \frac{7}{10}Mv^2$.
4. The work done by a force $\vec{F} = (2\hat{i} + 3\hat{j})\text{ N}$ in displacing a particle from $(r_1) = (\hat{i} - \hat{j})\text{ m}$ to $(r_2) = (3\hat{i} + 2\hat{j})\text{ m}$ is:
5. At what speed does the relativistic kinetic energy of a particle equal its rest mass energy?
Explanation: Given $KE = mc^2$, so $(\gamma - 1)mc^2 = mc^2 \implies \gamma = 2$. Since $\frac{1}{\sqrt{1-v^2/c^2}} = 2$, squaring gives $1 - v^2/c^2 = 1/4 \implies v^2/c^2 = 3/4 \implies v = \frac{\sqrt{3}}{2}c \approx 0.866c$.
6. A particle moves under the influence of a force such that its position as a function of time is $x(t) = t^3$. The instantaneous power delivered to the particle at $t = 2\text{ s}$ if its mass is $1\text{ kg}$ is:
Explanation: Position $x = t^3$, velocity $v = \frac{dx}{dt} = 3t^2$, acceleration $a = \frac{dv}{dt} = 6t$. Force $F = ma = 1 \times 6t = 6t$. Power $P = F \cdot v = (6t)(3t^2) = 18t^3$. At $t = 2\text{ s}$, $P = 18(2^3) = 18 \times 8 = 144\text{ W}$? Wait, let's check via $\frac{d(KE)}{dt}$: $KE = \frac{1}{2}mv^2 = \frac{1}{2}(1)(3t^2)^2 = \frac{9}{2}t^4$. Then $P = \frac{d(KE)}{dt} = \frac{9}{2}(4t^3) = 18t^3$. At $t=2$, $18(8) = 144\text{ W}$. Let's adjust options: Correct option index 1 was $72$, let's put $144\text{ W}$ in option index 1 and fix options.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. A bomb of mass $m$ at rest explodes into two fragments of masses $m_1$ and $m_2$. The ratio of their kinetic energies ($KE_1 / KE_2$) is:
Explanation: Since total initial momentum is zero, magnitudes of momenta of fragments are equal ($p_1 = p_2$). Kinetic energy $KE = \frac{p^2}{2m}$, so $\frac{KE_1}{KE_2} = \frac{p_1^2 / 2m_1}{p_2^2 / 2m_2} = \frac{m_2}{m_1}$.
2. The kinetic energy $KE$ of a particle moving along a straight line depends on distance $x$ as $KE = ax^2$ (where $a$ is a constant). The net force acting on the particle is:
Explanation: By work-energy theorem, $W_{net} = \Delta KE$. For infinitesimal displacement $dx$, $F \cdot dx = d(KE) = d(ax^2) = 2ax \, dx$, so $F = 2ax$, which is proportional to $x$.
3. A ring, a disc, and a solid sphere of equal mass and radius roll down an inclined plane from the same height without slipping. Which reaches the bottom with the highest kinetic energy?
Explanation: Since all bodies start from the same height with zero initial kinetic energy, their initial potential energy ($mgh$) is the same. By conservation of mechanical energy, all of them will have the exact same total kinetic energy at the bottom.
4. If the velocity of a particle is doubled, what happens to its kinetic energy?
Explanation: Kinetic energy is proportional to the square of velocity ($KE \propto v^2$). Doubling $v$ multiplies $KE$ by $2^2 = 4$.
5. Which of the following frames of reference is required for the standard work-energy theorem ($W_{net} = \Delta KE$) to hold without pseudo-forces?
Explanation: The standard work-energy theorem is formulated and holds directly in inertial reference frames.
6. What is the kinetic energy of a system of two particles of masses $1\text{ kg}$ and $3\text{ kg}$ moving with velocities $4\text{ m/s}$ and $2\text{ m/s}$ respectively, calculated via centre of mass?
Explanation: The total kinetic energy of any system of particles is rigorously separated into the translational kinetic energy of the centre of mass and internal kinetic energy relative to the centre of mass.
7. A particle of mass $m$ is moving in a circular path of radius $r$ with constant speed $v$. What is the net work done by the centripetal force in one complete revolution?
Explanation: Centripetal force is always perpendicular to the instantaneous velocity (and displacement vector), so the work done is zero ($W = \vec{F} \cdot \vec{d} = 0$). By work-energy theorem, speed and kinetic energy remain constant.
8. The dimensions of momentum squared divided by mass ($p^2/m$) are identical to the dimensions of:
Explanation: $p^2/m$ represents kinetic energy, which has the dimensions of energy ($[M L^2 T^{-2}]$).
9. If a constant force $F$ acts on a particle of mass $m$ initially at rest, the kinetic energy gained by the particle after travelling a distance $d$ is proportional to:
Explanation: By work-energy theorem, $W_{net} = \Delta KE \implies F \cdot d = KE_{final} - 0$, so $KE_{final} = Fd$, meaning kinetic energy is directly proportional to distance $d$.
10. In relativistic mechanics, as the velocity $v$ of a particle approaches the speed of light $c$, its kinetic energy:
Explanation: Since $KE = (\gamma - 1)mc^2$ and as $v \to c$, the Lorentz factor $\gamma \to \infty$, the kinetic energy approaches infinity.