Introduction and The Scalar Product

In physics, the terms 'work', 'energy', and 'power' have precise scientific meanings that may differ from their everyday usage. Energy is defined as the capacity to do work. Work is related to the force applied to an object and the displacement it undergoes. Power is the rate at which work is done or energy is transferred.

To quantitatively understand these concepts, especially work, we first need a mathematical tool for multiplying two vectors to get a scalar result. This tool is the scalar product or dot product.

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The Scalar Product (Dot Product)

The scalar product of two vectors A and B, denoted as $\textbf{A} \cdot \textbf{B}$, is a scalar quantity defined as:

$\textbf{A} \cdot \textbf{B} = |\textbf{A}| |\textbf{B}| \cos \theta = AB \cos \theta$

where $\theta$ is the angle between the two vectors.

Geometric Interpretation

The dot product can be interpreted as the product of the magnitude of one vector and the magnitude of the projection (component) of the second vector along the direction of the first.

$\textbf{A} \cdot \textbf{B} = A (B \cos \theta) = B (A \cos \theta)$

Properties of the Scalar Product

• Commutative Law: It does not matter in which order the vectors are multiplied.

$\textbf{A} \cdot \textbf{B} = \textbf{B} \cdot \textbf{A}$

• Distributive Law: The dot product is distributive over vector addition.

$\textbf{A} \cdot (\textbf{B} + \textbf{C}) = \textbf{A} \cdot \textbf{B} + \textbf{A} \cdot \textbf{C}$

Scalar Product in Component Form

For the orthogonal unit vectors $\hat{\textbf{i}}$, $\hat{\textbf{j}}$, and $\hat{\textbf{k}}$:

$\hat{\textbf{i}} \cdot \hat{\textbf{i}} = \hat{\textbf{j}} \cdot \hat{\textbf{j}} = \hat{\textbf{k}} \cdot \hat{\textbf{k}} = (1)(1)\cos 0^\circ = 1$

$\hat{\textbf{i}} \cdot \hat{\textbf{j}} = \hat{\textbf{j}} \cdot \hat{\textbf{k}} = \hat{\textbf{k}} \cdot \hat{\textbf{i}} = (1)(1)\cos 90^\circ = 0$

Given two vectors $\textbf{A} = A_x\hat{\textbf{i}} + A_y\hat{\textbf{j}} + A_z\hat{\textbf{k}}$ and $\textbf{B} = B_x\hat{\textbf{i}} + B_y\hat{\textbf{j}} + B_z\hat{\textbf{k}}$, their scalar product is:

$\textbf{A} \cdot \textbf{B} = A_x B_x + A_y B_y + A_z B_z$

From this, we can find the magnitude of a vector from its dot product with itself:

$\textbf{A} \cdot \textbf{A} = A_x A_x + A_y A_y + A_z A_z = A_x^2 + A_y^2 + A_z^2$

Since $\textbf{A} \cdot \textbf{A} = A A \cos 0^\circ = A^2$, we have $A^2 = A_x^2 + A_y^2 + A_z^2$.

A key property is that if two non-zero vectors A and B are perpendicular, their dot product is zero, since $\cos 90^\circ = 0$.

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Notions of Work and Kinetic Energy: The Work-Energy Theorem

The concepts of work and kinetic energy are deeply connected through a fundamental principle known as the work-energy theorem.

We start with the kinematic equation for motion with constant acceleration: $v^2 - u^2 = 2as$. Multiplying both sides by $m/2$ (where $m$ is the mass) gives:

$\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = m(as)$

Using Newton's second law, $F=ma$, we get:

$\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = Fs$

This can be generalized to three dimensions using the scalar product:

$\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = m\textbf{a} \cdot \textbf{d} = \textbf{F} \cdot \textbf{d}$

Defining Work and Kinetic Energy

• The quantity $\frac{1}{2}mv^2$ is defined as the Kinetic Energy (K) of the body. It is the energy a body possesses by virtue of its motion.
• The quantity $\textbf{F} \cdot \textbf{d}$ is defined as the Work (W) done by the force F on the body over a displacement d.

The Work-Energy (WE) Theorem

Using these definitions, the equation becomes:

$K_f - K_i = W_{net}$

where $K_i$ and $K_f$ are the initial and final kinetic energies, and $W_{net}$ is the work done by the net force.

Work-Energy Theorem: The change in the kinetic energy of a particle is equal to the work done on it by the net force.

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Answer:

Work

The work done by a constant force F on an object that undergoes a displacement d is defined as the scalar product of the force and displacement vectors.

$W = \textbf{F} \cdot \textbf{d} = Fd \cos \theta$

where $\theta$ is the angle between the force and displacement vectors. Work can also be seen as the product of the magnitude of displacement and the component of the force along the direction of displacement ($F \cos\theta$).

Conditions for Zero Work

Work done is zero if:

1. The displacement is zero ($d=0$). E.g., a weightlifter holding a weight steady does no work on the weight.
2. The force is zero ($F=0$). E.g., a block sliding on a frictionless horizontal surface.
3. The force and displacement are mutually perpendicular ($\theta = 90^\circ$). E.g., the gravitational force does no work on a satellite in a perfectly circular orbit, as the force is always perpendicular to the instantaneous velocity (displacement).

Positive and Negative Work

• Work is positive if the angle $\theta$ is acute ($0^\circ \le \theta < 90^\circ$). The force has a component in the direction of motion, aiding it.
• Work is negative if the angle $\theta$ is obtuse ($90^\circ < \theta \le 180^\circ$). The force has a component opposite to the direction of motion, opposing it. Work done by friction is a common example of negative work.

Units and Dimensions of Work

The dimensional formula for work is $[ML^2T^{-2}]$.

The SI unit of work is the joule (J). $1 \text{ J} = 1 \text{ N m}$.

Unit Name	Symbol	Value in J
erg	erg	$10^{-7}$ J
calorie	cal	$4.186$ J
kilowatt-hour	kWh	$3.6 \times 10^6$ J
electron volt	eV	$1.6 \times 10^{-19}$ J

Kinetic Energy

Kinetic Energy (K) is the energy possessed by an object by virtue of its motion. It is a scalar quantity and is defined as:

$K = \frac{1}{2}mv^2$

where $m$ is the mass of the object and $v$ is its speed. Since mass and the square of speed are always non-negative, kinetic energy is always a positive scalar quantity.

The kinetic energy of an object quantifies the amount of work the object can do on another object before coming to rest. For instance, the kinetic energy of wind is used by windmills, and the kinetic energy of flowing water is used in hydroelectric power plants.

Object	Car (60 km/h)	Running athlete (70 kg)	Bullet (50 g, 200 m/s)	Stone dropped from 10 m	Rain drop (1 g, 50 m/s)
Kinetic Energy (J)	$1.1 \times 10^5$	3000	1000	200	1.25

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Work Done by a Variable Force

In most real-world scenarios, the force acting on an object is not constant; it varies with position. To calculate the work done by a variable force $F(x)$ as an object moves from an initial position $x_i$ to a final position $x_f$, we must use integral calculus.

We can approximate the work done by dividing the path into many small displacements, $\Delta x$. Over each small displacement, the force is approximately constant. The total work is the sum of the work done over all these small segments:

$W \approx \sum_{x_i}^{x_f} F(x) \Delta x$

In the limit as the displacements approach zero ($\Delta x \to 0$), this sum becomes a definite integral:

$W = \int_{x_i}^{x_f} F(x) dx$

Graphical Interpretation

Graphically, the work done by a variable force is equal to the area under the force-displacement (F-x) curve between the initial and final positions.

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Answer:

The Work-Energy Theorem for a Variable Force

The work-energy theorem ($K_f - K_i = W_{net}$) holds true even when the net force is variable. We can prove this using calculus.

Proof

The time rate of change of kinetic energy is:

$\frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2}mv^2\right) = \frac{1}{2}m \cdot 2v \frac{dv}{dt} = mv \frac{dv}{dt}$

From Newton's second law, $F = ma = m\frac{dv}{dt}$. Substituting this, we get:

$\frac{dK}{dt} = F \cdot v$

Since velocity $v = \frac{dx}{dt}$, we can write $\frac{dK}{dt} = F \frac{dx}{dt}$.

Multiplying by $dt$, we get $dK = Fdx$.

Now, we integrate this expression from the initial state (position $x_i$, kinetic energy $K_i$) to the final state (position $x_f$, kinetic energy $K_f$):

$\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F(x) dx$

$K_f - K_i = W$

This proves that the change in kinetic energy is equal to the work done by the net force, even if the force is variable.

The Concept of Potential Energy

Potential energy (V) is the energy stored in an object by virtue of its position or configuration. It represents the 'potential' to do work or be converted into kinetic energy.

This concept is only applicable for a special class of forces called conservative forces. For such forces, the work done against them gets stored as potential energy. When the external constraints are removed, this stored energy is released, typically as kinetic energy.

Conservative Forces

A force is conservative if:

1. The work done by the force in moving an object between two points is independent of the path taken.
2. The work done by the force in a closed path (starting and ending at the same point) is zero.

Examples of conservative forces include the gravitational force and the spring force. Friction is a classic example of a non-conservative force.

Relation between Force and Potential Energy

For a conservative force $F(x)$, the potential energy $V(x)$ is defined such that the force is the negative gradient (in 1D, the negative derivative) of the potential energy.

$F(x) = -\frac{dV(x)}{dx}$

This implies that the change in potential energy, $\Delta V$, when an object moves from an initial position to a final position, is the negative of the work done by the conservative force.

$\Delta V = V_f - V_i = -W_c = -\int_{x_i}^{x_f} F_c(x) dx$

For example, for gravity near the Earth's surface, $F_g = -mg$ (upward is positive). The potential energy is $V(h) = mgh$, and indeed, $-\frac{d(mgh)}{dh} = -mg$.

The Conservation of Mechanical Energy

If only conservative forces are doing work on a system, its total mechanical energy is conserved.

Total Mechanical Energy (E) is the sum of the kinetic energy (K) and potential energy (V) of the system.

$E = K + V$

Proof of the Principle

From the work-energy theorem, the work done by the net conservative force, $W_c$, is equal to the change in kinetic energy:

$\Delta K = W_c$

From the definition of potential energy, the work done by the conservative force is also equal to the negative of the change in potential energy:

$W_c = -\Delta V$

Equating these two expressions:

$\Delta K = -\Delta V \implies \Delta K + \Delta V = 0 \implies \Delta(K+V) = 0$

Since the change in the total mechanical energy $(K+V)$ is zero, the quantity itself must be constant.

Principle of Conservation of Mechanical Energy: The total mechanical energy of a system is conserved if the forces doing work on it are conservative.

$K_i + V_i = K_f + V_f$

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Answer:

The Potential Energy of a Spring

The force exerted by an ideal spring is a conservative, variable force. It is described by Hooke's Law:

$F_s = -kx$

where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. The negative sign indicates that the force is a restoring force, always directed opposite to the displacement.

The work done by the spring force when it is stretched or compressed from $x=0$ to $x=x_m$ is:

$W_s = \int_{0}^{x_m} F_s dx = \int_{0}^{x_m} (-kx) dx = \left[-\frac{1}{2}kx^2\right]_{0}^{x_m} = -\frac{1}{2}kx_m^2$

Since the change in potential energy is the negative of the work done by the conservative force, we can define the elastic potential energy $V(x)$ of the spring. Taking $V(0)=0$ at the equilibrium position:

$V(x) = \frac{1}{2}kx^2$

For a block of mass $m$ attached to a spring, if there is no friction, the total mechanical energy is conserved:

$E = K + V = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{Constant}$

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Various Forms of Energy and The Law of Conservation of Energy

Mechanical energy is just one form of energy. Energy exists in many other forms, and they can be transformed into one another.

Other Forms of Energy

• Heat: The energy associated with the random motion of atoms and molecules within a system (internal energy). Work done by friction is converted into heat.
• Chemical Energy: Energy stored in the bonds of molecules. It is released in chemical reactions, such as combustion.
• Electrical Energy: Energy associated with the flow of electric charge (current).
• Nuclear Energy: Energy released from the nucleus of an atom during nuclear reactions like fission (splitting of nuclei) and fusion (joining of nuclei).

Mass-Energy Equivalence

Albert Einstein's theory of special relativity showed that mass and energy are equivalent and can be interconverted. They are related by the famous equation:

$E = mc^2$

where $c$ is the speed of light in vacuum. This equation implies that a small amount of mass can be converted into a vast amount of energy. This is the principle behind nuclear energy.

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The Principle of Conservation of Energy

When we account for all forms of energy, we arrive at one of the most fundamental laws of physics. If non-conservative forces are present, mechanical energy is not conserved, but it is transformed into other forms. The total energy, however, remains constant.

The Law of Conservation of Energy: Energy may be transformed from one form to another, but the total energy of an isolated system remains constant. Energy can neither be created, nor destroyed.

This principle is a unifying concept that connects all branches of science.

Power

Power is the rate at which work is done or energy is transferred.

Average Power

The average power, $P_{av}$, is the total work done divided by the total time taken.

$P_{av} = \frac{W}{t}$

Instantaneous Power

The instantaneous power, $P$, is the limiting value of the average power as the time interval approaches zero. It is the derivative of work with respect to time.

$P = \frac{dW}{dt}$

Since $dW = \textbf{F} \cdot d\textbf{r}$, we can also write power as:

$P = \frac{\textbf{F} \cdot d\textbf{r}}{dt} = \textbf{F} \cdot \left(\frac{d\textbf{r}}{dt}\right) = \textbf{F} \cdot \textbf{v}$

So, instantaneous power is the scalar product of the force and velocity vectors.

Units of Power

• The SI unit of power is the watt (W). $1 \text{ W} = 1 \text{ J/s}$.
• Another common unit is the horsepower (hp). $1 \text{ hp} \approx 746 \text{ W}$.
• The kilowatt-hour (kWh) is a unit of energy, not power. It is the energy consumed when 1 kilowatt of power is used for 1 hour. $1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$.

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Collisions

A collision is an event in which two or more bodies exert forces on each other in a relatively short time. The law of conservation of momentum is a key tool for analyzing all collisions.

Elastic and Inelastic Collisions

• In any collision, the total linear momentum of the system is always conserved (as long as the system is isolated).
• Elastic Collision: A collision in which the total kinetic energy of the system is also conserved. The objects deform during the collision but regain their original shape completely.
• Inelastic Collision: A collision in which the total kinetic energy of the system is not conserved. Some kinetic energy is converted into other forms, like heat, sound, or permanent deformation.
• Completely Inelastic Collision: A special case of inelastic collision where the objects stick together and move with a common velocity after the collision. The loss of kinetic energy is maximum in this case.

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Collisions in One Dimension (Head-on Collision)

Consider a mass $m_1$ with initial velocity $v_{1i}$ colliding with a mass $m_2$ at rest ($v_{2i}=0$). Let their final velocities be $v_{1f}$ and $v_{2f}$.

Completely Inelastic Collision

The objects stick together, so $v_{1f} = v_{2f} = v_f$.

By conservation of momentum: $m_1v_{1i} = (m_1+m_2)v_f$.

Final velocity: $v_f = \frac{m_1}{m_1+m_2}v_{1i}$.

Elastic Collision

We use both conservation of momentum and conservation of kinetic energy.

1. Momentum: $m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

2. Kinetic Energy: $\frac{1}{2}m_1v_{1i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

Solving these two equations simultaneously for the final velocities gives:

$v_{1f} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)v_{1i}$

$v_{2f} = \left(\frac{2m_1}{m_1 + m_2}\right)v_{1i}$

Special Case: Equal Masses ($m_1 = m_2$)

If the masses are equal, $v_{1f} = 0$ and $v_{2f} = v_{1i}$. The first mass stops, and the second mass moves off with the initial velocity of the first. This is commonly seen in billiards.

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Collisions in Two Dimensions (Glancing Collision)

For collisions that are not head-on, the final velocities are not along the initial line of motion. We must apply the conservation of momentum in vector form, which means conserving it along perpendicular axes separately.

Let $m_2$ be initially at rest, and $m_1$ have initial velocity $v_{1i}$ along the x-axis. After the collision, $m_1$ moves with velocity $v_{1f}$ at an angle $\theta_1$ and $m_2$ moves with velocity $v_{2f}$ at an angle $\theta_2$.

Conservation of momentum gives:

• x-component: $m_1v_{1i} = m_1v_{1f}\cos\theta_1 + m_2v_{2f}\cos\theta_2$
• y-component: $0 = m_1v_{1f}\sin\theta_1 - m_2v_{2f}\sin\theta_2$

If the collision is elastic, we also have the conservation of kinetic energy: $\frac{1}{2}m_1v_{1i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$.

An interesting result occurs for a glancing elastic collision between two equal masses when one is initially at rest: the two masses will move off at right angles to each other ($\theta_1 + \theta_2 = 90^\circ$).

Power Consumption in Walking

While the physiological feeling of "work" (getting tired) does not always align with the physics definition of mechanical work, we can estimate the mechanical work and power expended in common human activities. The table below gives some approximate values for the power consumed by a 60 kg adult in various activities.

Activity	Power (W)
Sleeping	75
Sitting	120
Bicycling	500
Swimming	600
Running (15 km/h)	1200

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Estimating the Mechanical Work in Walking

We can estimate the mechanical work done during walking by using the work-energy theorem. This allows us to bypass the difficult task of directly calculating the complex muscular forces involved. To do this, we make a simplified model with the following assumptions:

• (a) The primary work done is due to the repeated acceleration and deceleration of the legs with each stride.
• (b) Air resistance is neglected.
• (c) The small amount of work done to lift the legs against gravity is neglected.
• (d) Other movements, like the swinging of hands, are ignored.

In each stride, one leg is accelerated from rest to a speed approximately equal to the walking speed ($v_0$), and then decelerated back to rest. The other leg does the same in the next part of the stride. The work-energy theorem can be applied to this change in kinetic energy.

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Applying the Work-Energy Theorem and Calculating Power

Let $m_l$ be the mass of one leg.

According to the work-energy theorem, the work done by the leg muscles to accelerate the leg from rest to speed $v_0$ is equal to the change in its kinetic energy:

$W_{accel} = K_f - K_i = \frac{1}{2}m_l v_0^2 - 0 = \frac{1}{2}m_l v_0^2$

Similarly, another set of muscles does work to decelerate the leg from speed $v_0$ back to rest:

$W_{decel} = K_f - K_i = 0 - \frac{1}{2}m_l v_0^2 = -\frac{1}{2}m_l v_0^2$

The energy expended by the body for this deceleration is the magnitude of this work, which is $\frac{1}{2}m_l v_0^2$. Therefore, the total energy expended (work done) by one leg during its full motion in a stride is the sum of the energy for acceleration and deceleration:

$W_{one\_leg} = \frac{1}{2}m_l v_0^2 + \frac{1}{2}m_l v_0^2 = m_l v_0^2$

Since both legs perform this action during a full cycle (one stride for each leg), the total work done for both legs in one stride cycle is:

$W_s = 2(m_l v_0^2)$

Sample Calculation

Let's assume the mass of a leg, $m_l = 10$ kg.

Let the person be running slowly at a speed of 10 km/h. First, convert this speed to m/s:

$v_0 = 10 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} \approx 2.78 \text{ m/s} \approx 3 \text{ m/s}$.

The work done per stride cycle is:

$W_s = 2 \times (10 \text{ kg}) \times (3 \text{ m/s})^2 = 20 \times 9 = 180 \text{ J per stride cycle}$.

Now, let's calculate the power. If we assume a stride length of 2 m, the number of strides per second is:

Strides per second = $\frac{\text{Speed}}{\text{Stride Length}} = \frac{3 \text{ m/s}}{2 \text{ m/stride}} = 1.5 \text{ strides/s}$.

The power expended is the work done per second:

Power = $W_s \times (\text{strides per second}) = 180 \text{ J/stride} \times 1.5 \text{ strides/s} = 270 \text{ J/s} = 270 \text{ W}$.

This is a lower-end estimate, as we have ignored other energy loss mechanisms. This example highlights the advantage of using the work-energy theorem to analyze complex biological systems where forces are difficult to measure directly. It also helps us appreciate the efficiency of the wheel, which permits smooth motion without the continuous starting and stopping (acceleration and deceleration) inherent in walking and running.

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Additional Exercises

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