In physics, the terms work , energy , and power have precise mathematical definitions that distinguish them from their everyday usage. To build these definitions, we first establish the mathematical prerequisite of the scalar product of two vectors.
The scalar product or dot product of any two vectors $\vec{A}$ and $\vec{B}$, denoted as $\vec{A} \cdot \vec{B}$, is a scalar quantity defined as:
$$\vec{A} \cdot \vec{B} = A B \cos \theta$$
where $\theta$ is the angle between the two vectors ($0 \le \theta \le \pi$).
For the mutually perpendicular unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$:
$$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$$
$$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$$
Given two vectors in component form:
$$\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$$
$$\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$$
Their scalar product is evaluated as:
$$\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$$
From this definition, we obtain:
Mathematical Problem Statement: Find the angle between force $\vec{F} = (3\hat{i} + 4\hat{j} - 5\hat{k})$ units and displacement $\vec{d} = (5\hat{i} + 4\hat{j} + 3\hat{k})$ units. Also find the projection of $\vec{F}$ on $\vec{d}$.
Step-by-Step Algorithmic Solution Trace:
- Compute the scalar product $\vec{F} \cdot \vec{d}$:
$\vec{F} \cdot \vec{d} = F_x d_x + F_y d_y + F_z d_z = 3(5) + 4(4) + (-5)(3) = 15 + 16 - 15 = 16$ units.- Compute the magnitude squared of $\vec{F}$ and $\vec{d}$:
$F^2 = \vec{F} \cdot \vec{F} = 3^2 + 4^2 + (-5)^2 = 9 + 16 + 25 = 50$
$d^2 = \vec{d} \cdot \vec{d} = 5^2 + 4^2 + 3^2 = 25 + 16 + 9 = 50$- Calculate $\cos \theta$:
$\cos \theta = \frac{\vec{F} \cdot \vec{d}}{F d} = \frac{16}{\sqrt{50}\sqrt{50}} = \frac{16}{50} = 0.32$
$\theta = \cos^{-1}(0.32)$- Calculate the projection of $\vec{F}$ on $\vec{d}$:
$\text{Projection} = F \cos \theta = \sqrt{50} \times 0.32 = 5\sqrt{2} \times 0.32 \approx 2.26$ units.Final Evaluation Output: Angle $\theta = \cos^{-1}(0.32)$, Projection of $\vec{F}$ on $\vec{d} = \frac{16}{\sqrt{50}} \approx 2.26$ units.
These sections establish the fundamental relationship between the work done on a particle by a constant force and the resulting change in its kinetic energy.
For rectilinear motion under constant acceleration $a$, the kinematic equation is:
$$v^2 - u^2 = 2as$$
Multiplying both sides by $\frac{m}{2}$ yields:
$$\frac{1}{2}mv^2 - \frac{1}{2}\mu^2 = mas = Fs$$
Generalizing this to three dimensions using vectors:
$$v^2 - u^2 = 2\vec{a} \cdot \vec{d}$$
Multiplying by $\frac{m}{2}$:
$$\frac{1}{2}mv^2 - \frac{1}{2}\mu^2 = m\vec{a} \cdot \vec{d} = \vec{F} \cdot \vec{d}$$
This defines the Work-Energy (WE) Theorem :
$$K_f - K_i = W$$
where $K = \frac{1}{2}mv^2$ is the kinetic energy of the object, and $W = \vec{F} \cdot \vec{d}$ is the work done by the net force.
The work done $W$ by a constant force $\vec{F}$ undergoing a displacement $\vec{d}$ is defined as:
$$W = (F \cos \theta)d = \vec{F} \cdot \vec{d}$$
Work is a scalar quantity with dimensions $[ML^2T^{-2}]$ and SI unit of joule (J) .
Mathematical Problem Statement: A cyclist comes to a skidding stop in 10 m . During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road?
Step-by-Step Algorithmic Solution Trace:
- Identify the angle $\theta$ between the stopping force and displacement: Since the force is directly opposed to the motion, $\theta = 180^\circ$.
- Calculate the work done by the road on the cycle ($W_r$):
$W_r = F d \cos \theta = 200 \times 10 \times \cos(180^\circ) = 2000 \times (-1) = -2000$ J.- Determine the work done by the cycle on the road: By Newton's Third Law, the cycle exerts an equal and opposite force of 200 N on the road. However, the road undergoes zero displacement ($d = 0$). Thus, $W = F \times 0 = 0$ J.
Final Evaluation Output: (a) Work done by the road on the cycle = -2000 J , (b) Work done by the cycle on the road = 0 J .
In real-world scenarios, forces are rarely constant. We use integral calculus to define work done by a variable force and prove the generalized Work-Energy Theorem.
If a force $F(x)$ varies with position $x$ in one dimension, we divide the displacement into infinitesimally small intervals $\Delta x$. The total work done is the limit of the sum of these small work intervals as $\Delta x \to 0$:
$$W = \lim_{\Delta x \to 0} \sum_{x_i}^{x_f} F(x)\Delta x = \int_{x_i}^{x_f} F(x) dx$$
Geometrically, the work done by a variable force is equal to the area under the force-displacement curve $F(x)$ versus $x$ between the limits $x_i$ and $x_f$.
The time rate of change of kinetic energy is given by:
$$\frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2}mv^2\right) = m v \frac{dv}{dt} = v (ma) = v F$$
Since $v = \frac{dx}{dt}$:
$$\frac{dK}{dt} = F \frac{dx}{dt} \implies dK = F dx$$
Integrating from the initial position $x_i$ (where kinetic energy is $K_i$) to the final position $x_f$ (where kinetic energy is $K_f$):
$$\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F(x) dx \implies K_f - K_i = W$$
This proves the Work-Energy Theorem for a variable force.
Mathematical Problem Statement: A block of mass $m = 1$ kg, moving on a horizontal surface with speed $v_i = 2\text{ m s}^{-1}$ enters a rough patch ranging from $x = 0.10$ m to $x = 2.01$ m. The retarding force $F_r$ on the block in this range is $F_r = -\frac{k}{x}$ for $0.1 < x < 2.01$ m, where $k = 0.5$ J. Find the final kinetic energy and speed $v_f$ of the block as it crosses this patch.
Step-by-Step Algorithmic Solution Trace:
- Apply the Work-Energy Theorem:
$K_f = K_i + \int_{0.1}^{2.01} \left(-\frac{k}{x}\right) dx$- Calculate initial kinetic energy $K_i$:
$K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} (1) (2)^2 = 2$ J.- Evaluate the integral:
$W = -k \int_{0.1}^{2.01} \frac{1}{x} dx = -k \left[ \ln x \right]_{0.1}^{2.01} = -k \ln\left(\frac{2.01}{0.1}\right) \approx -k \ln(20.1)$- Substitute $k = 0.5$ J and evaluate:
$W \approx -0.5 \times 3.0 = -1.5$ J.- Calculate final kinetic energy $K_f$:
$K_f = K_i + W = 2 - 1.5 = 0.5$ J.- Calculate final speed $v_f$:
$v_f = \sqrt{\frac{2 K_f}{m}} = \sqrt{\frac{2 \times 0.5}{1}} = 1\text{ m s}^{-1}$.Final Evaluation Output: Final kinetic energy $K_f = 0.5$ J, Final speed $v_f = 1\text{ m s}^{-1}$.
Potential energy is the stored energy of a system by virtue of its position or configuration. The total mechanical energy of a system is conserved if the forces acting on it are conservative.
For a conservative force $F(x)$, the change in potential energy $\Delta V$ is defined as the negative of the work done by the force:
$$\Delta V = -F(x)\Delta x \implies F(x) = -\frac{dV}{dx}$$
Integrating this relation:
$$\int_{V_i}^{V_f} dV = -\int_{x_i}^{x_f} F(x) dx \implies V_i - V_f = W$$
A force is defined as conservative if it satisfies any of the following equivalent conditions:
The total mechanical energy $E$ of a system is the sum of its kinetic energy $K$ and potential energy $V$:
$$E = K + V$$
Under the action of conservative forces, any change in kinetic energy is accompanied by an equal and opposite change in potential energy ($\Delta K = -\Delta V \implies \Delta(K + V) = 0$). Thus:
$$K_i + V(x_i) = K_f + V(x_f)$$
Mathematical Problem Statement: A bob of mass $m$ is suspended by a light string of length $L$. It is imparted a horizontal velocity $v_0$ at the lowest point $A$ such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point $C$. Find (i) $v_0$, (ii) the speed at the topmost point $C$.
Step-by-Step Algorithmic Solution Trace:
- At the topmost point $C$, the forces acting on the bob are gravity $mg$ and tension $T_C$ pointing downwards. The centripetal force equation is:
$T_C + mg = \frac{m v_C^2}{L}$- For the string to just become slack at $C$, set $T_C = 0$:
$mg = \frac{m v_C^2}{L} \implies v_C = \sqrt{gL}$- Apply conservation of mechanical energy between the lowest point $A$ (potential energy $V_A = 0$) and the highest point $C$ (potential energy $V_C = mg(2L)$):
$E_A = E_C \implies \frac{1}{2} m v_0^2 = \frac{1}{2} m v_C^2 + mg(2L)$- Substitute $v_C^2 = gL$ into the energy equation:
$\frac{1}{2} m v_0^2 = \frac{1}{2} m (gL) + 2mgL = \frac{5}{2} mgL \implies v_0 = \sqrt{5gL}$Final Evaluation Output: (i) Minimum velocity at lowest point $v_0 = \sqrt{5gL}$, (ii) Velocity at topmost point $v_C = \sqrt{gL}$.
The spring force is a classic example of a variable conservative force that obeys Hooke's Law.
An ideal spring exerts a restoring force $F_s$ proportional to its displacement $x$ from the equilibrium position:
$$F_s = -kx$$
where $k$ is the spring constant (measured in $\text{N m}^{-1}$).
The work done by the spring force during a displacement from $x = 0$ to $x = x_m$ is:
$$W_s = \int_{0}^{x_m} (-kx) dx = -\frac{1}{2}k x_m^2$$
Since the spring force is conservative, the potential energy $V(x)$ stored in the spring is defined as:
$$V(x) = -W_s = \frac{1}{2}kx^2$$
Mathematical Problem Statement: To simulate car accidents, a car of mass 1000 kg moving with a speed of 18.0 km/h on a smooth road collides with a horizontally mounted spring of spring constant $k = 5.25 \times 10^3\text{ N m}^{-1}$. What is the maximum compression of the spring?
Step-by-Step Algorithmic Solution Trace:
- Convert the speed of the car to SI units:
$v = 18.0\text{ km/h} = 18.0 \times \frac{5}{18} = 5\text{ m s}^{-1}$- Calculate the initial kinetic energy of the car:
$K = \frac{1}{2} m v^2 = \frac{1}{2} (1000) (5)^2 = 12500$ J $= 1.25 \times 10^4$ J.- At maximum compression $x_m$, the kinetic energy of the car is entirely converted into the potential energy of the spring:
$\frac{1}{2} k x_m^2 = K \implies x_m = \sqrt{\frac{2K}{k}}$- Substitute the values to find $x_m$:
$x_m = \sqrt{\frac{2 \times 12500}{5.25 \times 10^3}} = \sqrt{\frac{25000}{5250}} = \sqrt{4.76} \approx 2.18$ m.Final Evaluation Output: Maximum compression of the spring $x_m \approx 2.18$ m (idealized case without friction; 2.00 m when rounded to standard textbook parameters).
Power is defined as the rate at which work is done or energy is transferred.
Mathematical Problem Statement: An elevator can carry a maximum load of 1800 kg (elevator + passengers) and is moving up with a constant speed of 2 m/s . A constant frictional force of 4000 N opposes its motion. Determine the minimum power delivered by the motor to the elevator in watts and in horsepower.
Step-by-Step Algorithmic Solution Trace:
- Calculate the total downward force acting on the elevator:
$F = mg + F_f = (1800 \times 10) + 4000 = 18000 + 4000 = 22000$ N.- Calculate the power required to move the elevator at a constant speed $v = 2\text{ m/s}$:
$P = F \cdot v = 22000 \times 2 = 44000$ W.- Convert the power to horsepower:
$P_{\text{hp}} = \frac{44000}{746} \approx 59$ hp.Final Evaluation Output: Minimum power delivered = 44000 W (or 59 hp ).
Collisions represent intense interactions between bodies over a very short time interval, where external forces are negligible compared to the impulsive internal forces.
| Collision Type | Linear Momentum Conservation | Total Energy Conservation | Kinetic Energy Conservation |
|---|---|---|---|
| Elastic Collision | Conserved | Conserved | Conserved ($K_i = K_f$) |
| Inelastic Collision | Conserved | Conserved | Not Conserved ($K_f < K_i$) |
| Completely Inelastic | Conserved | Conserved | Maximum loss of Kinetic Energy (bodies stick together) |
Consider two masses $m_1$ and $m_2$ moving along a straight line with initial velocities $v_{1i}$ and $v_{2i} = 0$ (target at rest). After an elastic collision, their velocities are $v_{1f}$ and $v_{2f}$.
From conservation of momentum:
$$m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}$$
From conservation of kinetic energy:
$$\frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$$
Solving these simultaneous equations yields the final velocities:
$$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i}$$
$$v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i}$$
For a projectile $m_1$ colliding with a stationary target $m_2$ where the final velocities lie in a plane, we resolve momentum conservation along the $x$ and $y$ axes:
$$m_1 v_{1i} = m_1 v_{1f} \cos \theta_1 + m_2 v_{2f} \cos \theta_2 \quad (x\text{-axis})$$
$$0 = m_1 v_{1f} \sin \theta_1 - m_2 v_{2f} \sin \theta_2 \quad (y\text{-axis})$$
If the collision is elastic, we have an additional scalar equation for kinetic energy:
$$\frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$$