Introduction: Beyond Newtonian Dynamics
While Newtonian dynamics, centered on the vector concept of force ($\vec{F}=m\vec{a}$), provides a complete framework for classical mechanics, it is not always the most efficient or insightful approach. The scalar concepts of work and energy offer a powerful alternative perspective, particularly for complex systems. This framework simplifies problems by focusing on the initial and final states of a system, often bypassing the intricate details of the interactions that occur in between. Here, we develop these concepts from first principles using the language of vector calculus.
1. Work Done by a Force
1.1 General Definition: The Line Integral
In the most general case, a force $\vec{F}$ can vary in magnitude and direction, and the path of its point of application can be a curve in three-dimensional space. The work done by this force as the particle moves from point A to point B along a specific path $C$ is defined by the line integral:
Here, $d\vec{r} = dx\,\hat{i} + dy\,\hat{j} + dz\,\hat{k}$ is the infinitesimal displacement vector along the path $C$. The dot product $\vec{F} \cdot d\vec{r}$ isolates the component of the force that acts along the direction of displacement, which is the only component that contributes to the work done. If the path is parameterized by a variable, say $t$, from $t_A$ to $t_B$, where $\vec{r}(t)$ is the position vector, then the integral becomes:
1.2 Conservative and Non-Conservative Forces
A crucial distinction is made between two types of forces based on the path dependence of the work they do.
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A force is conservative if the work it does on a particle moving between two points is independent of the path taken. This is equivalent to stating that the work done over any closed loop is zero: $\oint \vec{F}_{c} \cdot d\vec{r} = 0$. Examples include gravity and the electrostatic force. Mathematically, a conservative force can always be expressed as the negative gradient of a scalar potential energy function, $U$.
$\vec{F}_{c} = -\nabla U$This property is mathematically equivalent to the force having zero curl: $\nabla \times \vec{F}_{c} = 0$.
- A force is non-conservative if the work it does depends on the path. Friction and air resistance are classic examples. The work done by these dissipative forces is always negative and depends on the total path length.
2. Kinetic Energy and the Work-Energy Theorem
2.1 The Derivation
The Work-Energy Theorem is a direct consequence of Newton's Second Law. Consider the net work done by the total force $\vec{F}_{net}$ on a particle of mass $m$.
We use the chain rule to transform the integral. Knowing that acceleration is the time derivative of velocity, $\vec{a} = \frac{d\vec{v}}{dt}$, and velocity is the time derivative of position, $\vec{v} = \frac{d\vec{r}}{dt}$ (which implies $d\vec{r} = \vec{v} dt$), we can substitute these into the work integral:
The $dt$ terms cancel, allowing us to change the variable of integration from position (via time) to velocity:
To evaluate this integral, we use the product rule for the dot product: $d(\vec{v} \cdot \vec{v}) = d\vec{v} \cdot \vec{v} + \vec{v} \cdot d\vec{v} = 2(\vec{v} \cdot d\vec{v})$. Therefore, $\vec{v} \cdot d\vec{v} = \frac{1}{2} d(\vec{v} \cdot \vec{v})$. Since $\vec{v} \cdot \vec{v} = v^2$, the integral becomes trivial:
We define the kinetic energy as $K = \frac{1}{2}mv^2$. The theorem thus states that the net work done on an object equals the change in its kinetic energy.
2.2 Relativistic Kinetic Energy
In Special Relativity, the total energy of a particle is $E = \gamma m c^2$, where $m$ is the rest mass and $\gamma = (1 - v^2/c^2)^{-1/2}$ is the Lorentz factor. The rest energy is $E_0 = mc^2$. The kinetic energy is the difference between the total energy and the rest energy.
For small velocities ($v \ll c$), the term $x = v^2/c^2$ is very small. We can use the binomial expansion for $\gamma = (1-x)^{-1/2} \approx 1 + (-\frac{1}{2})(-x) + O(x^2)$, which simplifies to $\gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2}$. Substituting this into the equation for $K_{rel}$ recovers the classical form:
3. Potential Energy and Conservation of Energy
3.1 Potential Energy Function
For any conservative force $\vec{F}_c$, we can define a scalar potential energy function $U(\vec{r})$ such that the work done by the force is equal to the negative change in potential energy.
Only differences in potential energy are physically meaningful; the zero point of potential energy can be chosen arbitrarily. The force is recovered from the potential via the negative gradient operator, $\nabla = \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z}$.
For a one-dimensional system, this simplifies to $F_x(x) = -\frac{dU}{dx}$. Points where $F_x=0$ are equilibrium points. The equilibrium is stable if $\frac{d^2U}{dx^2} > 0$ (a local minimum in $U$) and unstable if $\frac{d^2U}{dx^2} < 0$ (a local maximum in $U$).
3.2 The Law of Conservation of Energy
The net work can be split into work done by conservative forces ($W_c$) and work done by non-conservative forces ($W_{nc}$). From the Work-Energy Theorem:
Substituting $W_c = -\Delta U$, we get:
We define the total mechanical energy as $E_{mech} = K + U$. This yields the most general form of the law of energy conservation:
In an isolated system where only conservative forces do work ($W_{nc} = 0$), the total mechanical energy is conserved: $\Delta E_{mech} = 0$, or $K_i + U_i = K_f + U_f$.
4. Theoretical Examples
4.1 Classical: The Simple Pendulum
Consider a simple pendulum of mass $m$ and length $l$. The kinetic energy is $K = \frac{1}{2}mv^2 = \frac{1}{2}m(l\dot{\theta})^2$. The potential energy, setting $U=0$ at the lowest point, is $U = mgh = mgl(1-\cos\theta)$. If energy is conserved, the total energy $E$ is constant:
Differentiating with respect to time gives the equation of motion:
This demonstrates how the equation of motion can be derived elegantly from energy principles.
4.2 Modern Physics: The Morse Potential
The interaction between two atoms in a diatomic molecule can be approximated by the Morse potential, which accounts for bond stretching and dissociation.
Here, $r$ is the interatomic distance, $r_e$ is the equilibrium bond distance, $D_e$ is the well depth (dissociation energy), and $a$ controls the 'width' of the potential. The force between the atoms is $F(r) = -dU/dr$. For small displacements from equilibrium, a Taylor expansion of $U(r)$ around $r_e$ shows that the potential is approximately quadratic, corresponding to a simple harmonic oscillator, which explains the quantized vibrational energy levels of the molecule.
Practice Problems
Intermediate Level Problems
Problem 1:
A particle moves under the influence of a force $\vec{F} = (3x^2 \hat{i} + 4y \hat{j})$ N. Calculate the work done by this force as the particle moves from the origin (0,0) to the point (2,4) along the parabolic path $y=x^2$.
Problem 2:
A block of mass $m=5$ kg is released from rest at the top of an inclined plane of angle $\theta=30^\circ$ and height $h=10$ m. The coefficient of kinetic friction between the block and the plane is $\mu_k=0.2$. Using the work-energy theorem, find the speed of the block when it reaches the bottom of the incline.
Problem 3:
A potential energy function for a particle moving along the x-axis is given by $U(x) = 8x^2 - x^4$, where $U$ is in joules and $x$ is in meters. Determine (a) the points of stable and unstable equilibrium and (b) the force on the particle at $x=-1$ m.
Problem 4:
Derive the expression for the escape velocity from a planet of mass $M$ and radius $R$ by considering the conservation of mechanical energy. Calculate this velocity for Earth ($M \approx 5.97 \times 10^{24}$ kg, $R \approx 6.37 \times 10^6$ m).
Problem 5:
A mass $m$ is attached to a spring with spring constant $k$. The mass is pulled a distance $A$ from its equilibrium position and released from rest. Calculate the velocity of the mass as a function of its position $x$ using energy conservation.
Problem 6:
A proton (mass $m_p = 1.67 \times 10^{-27}$ kg) is accelerated from rest through a potential difference of $\Delta V = 2.5 \times 10^5$ V. What is its final kinetic energy (in Joules and eV) and its final speed? Is a relativistic treatment necessary? ($e = 1.602 \times 10^{-19}$ C).
Problem 7:
A bead of mass $m$ slides without friction on a vertical circular hoop of radius $R$. If it is displaced slightly from the top point, find its speed when it reaches the bottom point and the normal force exerted by the hoop at that point.
Advanced Level Problems
Problem 8:
A particle of mass $m$ is subject to a central force $\vec{F} = -k/r^3 \vec{r}$, where $k$ is a positive constant and $\vec{r}$ is the position vector from the origin. (a) Show that this force is conservative. (b) Find the potential energy function $U(r)$, assuming $U(\infty)=0$. (c) If the particle is projected from $r_0$ with a velocity $v_0$ perpendicular to $\vec{r}_0$, find the condition for which it will move in a circular orbit.
Problem 9:
A block of mass $m$ is placed on top of a larger block of mass $M$ which rests on a frictionless horizontal surface. A constant horizontal force $F$ is applied to the lower block. The coefficient of static friction between the blocks is $\mu_s$. Using work-energy considerations, find the maximum force $F_{max}$ that can be applied such that the top block does not slip. What is the work done by static friction on the top block in this case?
Problem 10:
A particle moves in the xy-plane with a velocity $\vec{v}(t) = (\alpha t)\hat{i} + (\beta t^2)\hat{j}$, where $\alpha$ and $\beta$ are constants. The particle starts at the origin at $t=0$. Find the net work done on the particle by the forces acting on it during the interval from $t=0$ to $t=T$.
Problem 11:
An electron with a kinetic energy of 1.0 MeV collides head-on with a stationary positron (same rest mass, opposite charge). They annihilate, creating two photons of equal energy that travel in opposite directions. Calculate the energy (in MeV) and wavelength of each photon.
Problem 12:
A small object of mass $m$ is released from rest at the top of a frictionless hemisphere of radius $R$. At what angle $\theta$ (measured from the vertical) does the object lose contact with the hemisphere?
Irodov-like Problems
Problem 13:
A uniform chain of length $L$ and mass $M$ lies on a frictionless horizontal table such that a small fraction of its length, $L_0$, hangs over the edge. The chain is released from rest. Find the velocity of the chain at the moment the last link leaves the table. You must solve this by setting up an integral based on the work-energy theorem, considering the work done by gravity on the center of mass of the hanging part.
Problem 14:
A particle of mass $m$ moves in a one-dimensional potential field $U(x) = U_0 \left( \frac{a^2}{x^2} - \frac{a}{x} \right)$, where $U_0$ and $a$ are positive constants. The particle has a total energy $E = -U_0/16$. Determine the turning points of the motion and calculate the maximum speed of the particle during its motion.
Problem 15:
A light inextensible thread is wound around a uniform solid cylinder of mass $M$ and radius $R$. The cylinder is placed on a rough horizontal plane with a coefficient of kinetic friction $\mu_k$. The free end of the thread is pulled with a constant horizontal force $F$. Assuming the cylinder rolls without slipping, find the kinetic energy of the cylinder after its center has moved a distance $S$. Now, consider the case where the force $F$ is large enough to cause slipping. Find the work done by friction and the total kinetic energy (translational + rotational) after the center has moved a distance $S$.
Solutions to Practice Problems
Intermediate Level Solutions
Solution 1:
The work done is $W = \int_C \vec{F} \cdot d\vec{r}$. We have $\vec{F} = 3x^2 \hat{i} + 4y \hat{j}$ and $d\vec{r} = dx \hat{i} + dy \hat{j}$.
The path is $y=x^2$, so $dy = 2x \, dx$. We substitute this into $d\vec{r}$ and $\vec{F}$:
Solution 2:
We use the work-energy theorem: $W_{net} = \Delta K = K_f - K_i$. Since the block starts from rest, $K_i=0$.
The net work is the sum of work done by gravity ($W_g$) and friction ($W_f$).
Solution 3:
(a) Equilibrium points occur where the force is zero. $F(x) = -dU/dx = 0$.
To check for stability, we examine the second derivative of $U(x)$.
- At $x=2$: $d^2U/dx^2 = 16 - 12(2^2) = -32 < 0 \implies$ Unstable equilibrium.
- At $x=-2$: $d^2U/dx^2 = 16 - 12(-2)^2) = -32 < 0 \implies$ Unstable equilibrium.
(b) The force at $x=-1$ m is:
Solution 4:
To escape the planet's gravity, the object must reach an infinite distance with at least zero kinetic energy. We set the final energy $E_f = K_f + U_f = 0 + 0 = 0$.
The initial state is at the planet's surface (radius $R$) with escape velocity $v_{esc}$.
Solution 5:
The total mechanical energy of the spring-mass system is conserved. $E = K+U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$.
Initially, at $x=A$, the mass is at rest ($v=0$). The total energy is purely potential:
Solution 6:
The work done by the electric field provides kinetic energy to the proton: $\Delta K = W = q\Delta V$.
To find the speed, we use the classical kinetic energy formula first.
Solution 7:
We use energy conservation. Let the bottom of the hoop be the reference for potential energy ($h=0$). The top is at $h=2R$.
Advanced Level Solutions
Solution 8:
(a) A force is conservative if its curl is zero, $\nabla \times \vec{F} = 0$. Given $\vec{F} = -k \frac{\vec{r}}{r^3} = -k\frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^2+y^2+z^2)^{3/2}}$. Let's check the z-component of the curl: $(\nabla \times \vec{F})_z = \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}$.
(b) Since the force is conservative, $F_r = -dU/dr$. The magnitude of the force is $|\vec{F}| = k/r^3 |\vec{r}| = k/r^2$. The force is attractive, so $F_r = -k/r^2$.
Solution 9:
The top block $m$ is accelerated by the static friction force $f_s$. The maximum acceleration it can have is $a_{max} = f_{s,max}/m = \mu_s N_m / m = \mu_s mg / m = \mu_s g$. For the blocks to move together without slipping, the whole system must have this maximum acceleration. Applying Newton's second law to the combined system $(M+m)$:
Solution 10:
The simplest way to solve this is using the Work-Energy Theorem, $W_{net} = \Delta K$. We just need to find the change in kinetic energy.
The initial velocity at $t=0$ is $\vec{v}(0) = 0$, so the initial kinetic energy $K_i=0$.
The final velocity at $t=T$ is:
Solution 11:
This problem requires conservation of total relativistic energy. The rest energy of an electron (or positron) is $E_0 = m_e c^2 \approx 0.511\,\text{MeV}$.
The total initial energy of the system is the sum of the energies of the electron and the positron.
The electron is moving, so its energy is its kinetic energy plus its rest energy: $E_{electron} = K_e + E_0$.
Solution 12:
Let $\theta$ be the angle from the vertical. Set potential energy $U=0$ at the initial height (top of the hemisphere). At an angle $\theta$, the object has dropped a vertical distance of $h=R(1-\cos\theta)$. By conservation of energy: $\Delta K = -\Delta U$.
Irodov-like Solutions
Solution 13:
We use the conservation of energy. Let the linear mass density be $\lambda = M/L$. Let the table surface be the zero potential energy reference ($U=0$).
Initial State: A length $L_0$ hangs. Its mass is $m_0 = \lambda L_0$. Its center of mass is at a distance $L_0/2$ below the table. The initial potential energy is $U_i = -m_0 g (L_0/2) = -(\lambda L_0)g(L_0/2) = -\frac{1}{2}\lambda g L_0^2$. The initial kinetic energy is $K_i=0$.
Solution 14:
Turning points occur where the kinetic energy is zero, so the total energy equals the potential energy: $E=U(x)$.
Solution 15:
Case 1: Rolls without slipping.
We use the work-energy theorem for rigid bodies: $W_{net} = \Delta K$. The net work is done only by the external force $F$, since the static friction does no work in pure rolling. $W_{net} = F \cdot S$.
The final kinetic energy has translational and rotational parts: $K_f = K_{trans} + K_{rot} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$.
For a solid cylinder, $I_{cm} = \frac{1}{2}MR^2$. For pure rolling, $v_{cm} = \omega R$.
Now kinetic friction acts, and $v_{cm} \neq \omega R$. The work done by friction is $W_f = -f_k S = -\mu_k N S = -\mu_k MgS$. The work-energy theorem for translation: $W_{net, trans} = \Delta K_{trans}$.