Introduction

The terms 'work', 'energy', and 'power' have specific, precise meanings in physics, which may differ from their everyday usage.

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In physics, Work is formally defined based on force and displacement.

Energy is closely related to the capacity to do work.

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

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To understand these concepts fully, particularly in multi-dimensional motion, a mathematical tool called the scalar product (or dot product) of two vectors is essential.

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The Scalar Product

The scalar product or dot product of any two vectors $\mathbf{A}$ and $\mathbf{B}$ is denoted by $\mathbf{A} \cdot \mathbf{B}$ and is defined as:

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

where $A$ and $B$ are the magnitudes of vectors $\mathbf{A}$ and $\mathbf{B}$ respectively, and $\theta$ is the angle between the two vectors.

The result of a scalar product is a scalar quantity (a number), even though the original quantities are vectors. It has magnitude but no direction.

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Geometric Interpretation:

• $\mathbf{A} \cdot \mathbf{B} = A (B \cos \theta)$: This is the product of the magnitude of $\mathbf{A}$ and the component of $\mathbf{B}$ along the direction of $\mathbf{A}$ ($B \cos \theta$ is the projection of $\mathbf{B}$ onto $\mathbf{A}$).
• $\mathbf{A} \cdot \mathbf{B} = B (A \cos \theta)$: This is the product of the magnitude of $\mathbf{B}$ and the component of $\mathbf{A}$ along the direction of $\mathbf{B}$ ($A \cos \theta$ is the projection of $\mathbf{A}$ onto $\mathbf{B}$).

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Properties of Scalar Product:

• Commutative Law: The order of multiplication does not matter. $$ \mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A} $$
• Distributive Law: Scalar product distributes over vector addition. $$ \mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = \mathbf{A} \cdot \mathbf{B} + \mathbf{A} \cdot \mathbf{C} $$
• Multiplication by a real number $\lambda$: $$ \mathbf{A} \cdot (\lambda \mathbf{B}) = \lambda (\mathbf{A} \cdot \mathbf{B}) $$

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Scalar Product in Component Form:

For unit vectors $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, $\hat{\mathbf{k}}$ along the x, y, z axes:

$$ \hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{k}} = 1 \quad (\text{since angle is } 0^\circ, \cos 0^\circ = 1) $$ $$ \hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{k}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{i}} = 0 \quad (\text{since angle is } 90^\circ, \cos 90^\circ = 0) $$

If two vectors are given in component form $\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}}$ and $\mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}} + B_z \hat{\mathbf{k}}$, their scalar product is:

$$ \mathbf{A} \cdot \mathbf{B} = (A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}}) \cdot (B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}} + B_z \hat{\mathbf{k}}) $$

Using the distributive property and the dot products of unit vectors:

$$ \mathbf{A} \cdot \mathbf{B} = A_x B_x (\hat{\mathbf{i}} \cdot \hat{\mathbf{i}}) + A_x B_y (\hat{\mathbf{i}} \cdot \hat{\mathbf{j}}) + A_x B_z (\hat{\mathbf{i}} \cdot \hat{\mathbf{k}}) + \dots $$ $$ \mathbf{A} \cdot \mathbf{B} = A_x B_x (1) + A_x B_y (0) + A_x B_z (0) + A_y B_x (0) + A_y B_y (1) + A_y B_z (0) + A_z B_x (0) + A_z B_y (0) + A_z B_z (1) $$ $$ \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z $$

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Special Cases:

• The dot product of a vector with itself gives the square of its magnitude: $$ \mathbf{A} \cdot \mathbf{A} = A_x A_x + A_y A_y + A_z A_z = A_x^2 + A_y^2 + A_z^2 = A^2 $$ since $\theta = 0^\circ$ and $\cos 0^\circ = 1$.
• If two non-zero vectors $\mathbf{A}$ and $\mathbf{B}$ are perpendicular, their dot product is zero: $$ \mathbf{A} \cdot \mathbf{B} = AB \cos 90^\circ = AB(0) = 0 $$ This is because the component of one vector along the direction perpendicular to it is zero.

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Answer:

Notions Of Work And Kinetic Energy : The Work-Energy Theorem

In physics, the concepts of work and energy are formally linked through the Work-Energy Theorem.

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For motion under a constant force $\mathbf{F}$ causing a displacement $\mathbf{d}$, the work done is defined as the scalar product $\mathbf{F} \cdot \mathbf{d}$.

A relationship derived from kinematics for constant acceleration is $v^2 - u^2 = 2 \mathbf{a} \cdot \mathbf{d}$, where $\mathbf{u}$ and $\mathbf{v}$ are initial and final velocities, $\mathbf{a}$ is acceleration, and $\mathbf{d}$ is displacement.

Multiplying this equation by $m/2$ (where $m$ is mass):

$$ \frac{1}{2}m v^2 - \frac{1}{2}m u^2 = m \mathbf{a} \cdot \mathbf{d} $$

From Newton's Second Law, $\mathbf{F} = m\mathbf{a}$. Substituting this gives:

$$ \frac{1}{2}m v^2 - \frac{1}{2}m u^2 = \mathbf{F} \cdot \mathbf{d} $$

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This equation motivates the definitions of Kinetic Energy and Work:

• Kinetic Energy (K): The quantity $\frac{1}{2}mv^2$ is defined as the kinetic energy of an object of mass $m$ moving with speed $v$. It is energy associated with motion. ($K = \frac{1}{2}m v^2$).
• Work (W): The quantity $\mathbf{F} \cdot \mathbf{d}$ is defined as the work done by the force $\mathbf{F}$ on the object over the displacement $\mathbf{d}$. ($W = \mathbf{F} \cdot \mathbf{d}$).

With these definitions, the equation becomes the Work-Energy (WE) Theorem:

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

$$ K_f - K_i = W_{net} $$

where $K_i = \frac{1}{2}mu^2$ is the initial kinetic energy and $K_f = \frac{1}{2}mv^2$ is the final kinetic energy.

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The WE theorem is a powerful tool because it relates the change in an object's motion (specifically, its kinetic energy) to the work done on it by forces, without necessarily needing to analyze the acceleration at every instant (unlike Newton's Second Law directly).

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Answer:

Work

In physics, work (W) done by a constant force $\mathbf{F}$ on an object over a displacement $\mathbf{d}$ is defined as the scalar product of the force and displacement vectors:

$$ W = \mathbf{F} \cdot \mathbf{d} $$

Alternatively, it is the product of the component of the force in the direction of the displacement and the magnitude of the displacement:

$$ W = (F \cos \theta) d $$

where $\theta$ is the angle between $\mathbf{F}$ and $\mathbf{d}$.

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Work is a scalar quantity. It can be positive, negative, or zero.

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

Dimensions of work are $[ML^2T^{-2}]$, which are the same as energy.

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Conditions for zero work done:

• If the displacement is zero ($\mathbf{d}=\mathbf{0}$), even if a force is applied (e.g., pushing a stationary wall, holding a heavy object without moving).
• If the force is zero ($\mathbf{F}=\mathbf{0}$), even if there is displacement (this is an idealized case; in reality, net force might be zero due to balanced forces, allowing uniform motion).
• If the force and displacement are mutually perpendicular ($\theta = 90^\circ$), since $\cos 90^\circ = 0$. (e.g., gravitational force on an object moving horizontally, centripetal force on an object in uniform circular motion).

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Sign of Work Done:

• Positive Work: If the force (or its component) is in the same direction as the displacement ($0^\circ \le \theta < 90^\circ$, $\cos \theta > 0$). The force tends to increase the object's speed.
• Negative Work: If the force (or its component) is in the opposite direction to the displacement ($90^\circ < \theta \le 180^\circ$, $\cos \theta < 0$). The force tends to decrease the object's speed. Frictional force and air resistance often do negative work on a moving object.
• Zero Work: If the force is perpendicular to the displacement ($\theta = 90^\circ$, $\cos \theta = 0$). The force does not change the object's speed (though it might change its direction, as in uniform circular motion).

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Alternative Units of Work/Energy (besides Joule):

Unit	Equivalent in Joules
erg (CGS unit of work)	$10^{-7}$ J
calorie (cal) (unit of heat)	$\approx 4.186$ J
electron volt (eV) (unit of energy in atomic/nuclear physics)	$1.602 \times 10^{-19}$ J
kilowatt hour (kWh) (unit of electrical energy)	$3.6 \times 10^6$ J
thermochemical calorie	4.184 J
British thermal unit (BTU)	$\approx 1055$ J

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Answer:

Kinetic Energy

Kinetic Energy (K) is the energy possessed by an object due to its motion.

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For an object of mass $m$ moving with velocity $\mathbf{v}$, its kinetic energy is defined as:

$$ K = \frac{1}{2} m |\mathbf{v}|^2 = \frac{1}{2} m v^2 $$

It can also be written as the scalar product $K = \frac{1}{2} \mathbf{p} \cdot \mathbf{v}$, where $\mathbf{p} = m\mathbf{v}$ is the momentum, or $K = \frac{|\mathbf{p}|^2}{2m}$.

Kinetic energy is a scalar quantity and is always non-negative ($m$ and $v^2$ are always positive or zero).

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The kinetic energy of an object represents the amount of work the object can do by virtue of its motion (e.g., a moving hammer can drive a nail, a moving car can crush something). It also represents the work done by the net force to bring the object from rest to its current velocity.

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Typical kinetic energies vary greatly depending on the mass and speed of the object.

Object	Speed	Kinetic Energy (approx)
Falling raindrop (1g)	Few m/s	$10^{-2}$ J
Air molecule ($\approx 10^{-26}$ kg) at room temperature	$\approx 500$ m/s	$10^{-21}$ J
Cricket ball (150g) bowled by a fast bowler	$\approx 40$ m/s	120 J
Bullet (50g)	$\approx 200$ m/s	1000 J
Running man (60kg)	$\approx 5$ m/s	750 J
Car (2000kg) at 60 km/h (16.7 m/s)	16.7 m/s	$2.8 \times 10^5$ J
Train (5 x $10^5$ kg) at 60 km/h	16.7 m/s	$7.0 \times 10^7$ J
Orbiting satellite (1000kg)	$\approx 8000$ m/s	$3.2 \times 10^{10}$ J

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Answer:

Work Done By A Variable Force

In many realistic situations, the force acting on an object is not constant but varies with position, time, or velocity. When the force is variable, the definition of work needs to account for this variation.

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Consider a force $F(x)$ acting on an object moving along the x-axis, where the force magnitude (and possibly direction) changes with position $x$. To calculate the work done by this variable force as the object moves from an initial position $x_i$ to a final position $x_f$, we can divide the total displacement into many small segments $\Delta x$.

Within a very small displacement $\Delta x$, the force $F(x)$ can be considered approximately constant. The work done during this small segment is $\Delta W \approx F(x) \Delta x$.

The total work done over the entire displacement from $x_i$ to $x_f$ is the sum of the work done over all small segments:

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

As the size of the segments $\Delta x$ approaches zero, the sum becomes an integral, and the approximation becomes exact. The work done by a variable force $F(x)$ is the definite integral of $F(x)$ with respect to $x$ from $x_i$ to $x_f$. Graphically, this is the area under the Force-Position curve.

$$ W = \lim_{\Delta x \to 0} \sum_{x_i}^{x_f} F(x) \Delta x = \int_{x_i}^{x_f} F(x) \, dx $$

For a force that varies in two or three dimensions ($\mathbf{F}(\mathbf{r})$), the work done over a path from initial position $\mathbf{r}_i$ to final position $\mathbf{r}_f$ is the line integral of the dot product of the force and the infinitesimal displacement vector $d\mathbf{r}$ along the path:

$$ W = \int_{\mathbf{r}_i}^{\mathbf{r}_f} \mathbf{F}(\mathbf{r}) \cdot d\mathbf{r} $$

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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 acting on a particle is variable.

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For one-dimensional motion, consider the rate of change of kinetic energy with time:

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

Using Newton's Second Law, $F = ma = m \frac{dv}{dt}$, so $m \frac{dv}{dt} = F$.

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

Rearranging, we get $dK = F \, dx$.

Integrate both sides from initial position $x_i$ to final position $x_f$ (corresponding to initial kinetic energy $K_i$ and final kinetic energy $K_f$):

$$ \int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F \, dx $$ $$ [K]_{K_i}^{K_f} = \int_{x_i}^{x_f} F \, dx $$ $$ K_f - K_i = \int_{x_i}^{x_f} F \, dx $$

The integral $\int_{x_i}^{x_f} F \, dx$ represents the total work done by the force $F$ over the displacement from $x_i$ to $x_f$.

So, $K_f - K_i = W$. This proves the Work-Energy Theorem for a variable force in one dimension.

The theorem also holds for variable forces in two and three dimensions, using the line integral definition of work: $\int_{\mathbf{r}_i}^{\mathbf{r}_f} \mathbf{F} \cdot d\mathbf{r}$.

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Limitations/Insights of the Work-Energy Theorem:

• It is an integral form of Newton's Second Law. While derived from it, it does not contain all the information of the Second Law. Newton's Second Law relates force and acceleration *at every instant*, while the WE theorem relates work done *over an interval* to the *change in kinetic energy over that interval*. Temporal (time) information is lost in the work integral.
• It is a scalar relation ($K$ and $W$ are scalars), while Newton's Second Law is a vector relation ($\mathbf{F}=m\mathbf{a}$). Directional information is integrated out.
• Despite limitations, it is very useful for solving problems, especially when force varies or when only initial/final speeds and total work are needed.

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Answer:

The Concept Of Potential Energy

Potential Energy (V) is the energy stored in a system due to the relative positions or configuration of its parts. It represents the potential or capacity to do work.

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The concept of potential energy is associated with a specific type of force called a conservative force.

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For a force $\mathbf{F}$ (in 1D, $F(x)$) to be conservative, there must exist a scalar potential energy function $V(x)$ such that the force is the negative derivative of this function with respect to position:

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

In three dimensions, this generalizes to $\mathbf{F}(\mathbf{r}) = -\nabla V(\mathbf{r})$, where $\nabla V$ is the gradient of the scalar function $V$.

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This definition implies that the work done by a conservative force as a particle moves from an initial position $x_i$ to a final position $x_f$ is given by:

$$ W = \int_{x_i}^{x_f} F(x) \, dx = \int_{x_i}^{x_f} \left(-\frac{dV}{dx}\right) \, dx = -[V(x)]_{x_i}^{x_f} = -(V(x_f) - V(x_i)) = V(x_i) - V(x_f) = -\Delta V $$

So, the work done by a conservative force is equal to the negative of the change in potential energy.

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Properties of Conservative Forces (equivalent definitions):

1. They can be derived from a potential energy function ($F = -dV/dx$).
2. The work done by the force on an object moving between two points depends only on the initial and final positions of the object, not on the specific path taken.
3. The work done by the force on an object moving along any closed path (returning to its starting point) is zero.

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Examples of Conservative Forces:

• Gravitational Force: Near the Earth's surface, the gravitational force on mass $m$ is approximately constant, $F_g = -mg$ (taking positive direction upwards). The potential energy function is $V(y) = mgy$, where $y$ is the height above a reference level (where $V=0$). $F_g = -\frac{d}{dy}(mgy) = -mg$. The work done by gravity is $W_g = \int_{y_i}^{y_f} (-mg) dy = -mg(y_f - y_i) = mgy_i - mgy_f$. This depends only on initial and final heights.
• Spring Force: The force exerted by an ideal spring is $F_s = -kx$, where $x$ is displacement from equilibrium. The potential energy function is $V(x) = \frac{1}{2}kx^2$, setting $V=0$ at $x=0$. $F_s = -\frac{d}{dx}(\frac{1}{2}kx^2) = -kx$. The work done by the spring force is $W_s = \int_{x_i}^{x_f} (-kx) dx = [-\frac{1}{2}kx^2]_{x_i}^{x_f} = -\frac{1}{2}kx_f^2 + \frac{1}{2}kx_i^2 = V(x_i) - V(x_f)$. This depends only on initial and final positions.

Examples of Non-Conservative Forces: Friction, air resistance, viscous drag. Work done by these forces depends on the path taken and is generally non-zero for a closed path. Mechanical energy is not conserved when these forces do work.

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The choice of the zero point for potential energy is arbitrary. It sets the reference level. However, once chosen, it must be used consistently throughout the problem. The *change* in potential energy ($\Delta V$) is physically meaningful and independent of the zero point choice.

Dimensions of potential energy are $[ML^2T^{-2}]$, same as work and kinetic energy. Unit is Joule (J).

Potential energy can be thought of as 'stored work' that can be released and converted into kinetic energy or other forms of energy when constraints are removed.

The Conservation Of Mechanical Energy

The principle of conservation of mechanical energy is a fundamental result that arises when only conservative forces do work on a system.

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Consider an object or a system on which only conservative forces are doing work. From the Work-Energy Theorem, the work done by the net force is equal to the change in kinetic energy:

$$ W_{net} = \Delta K $$

If all forces are conservative, the net force $\mathbf{F}_{net}$ is conservative, and the work done by the net force is $W_{net} = -\Delta V_{net}$, where $V_{net}$ is the total potential energy associated with the conservative forces.

Substituting this into the Work-Energy Theorem:

$$ -\Delta V_{net} = \Delta K $$ $$ \Delta K + \Delta V_{net} = 0 $$ $$ \Delta (K + V_{net}) = 0 $$

This means the sum of the kinetic energy and the total potential energy remains constant:

$$ K + V_{net} = \text{Constant} $$

The sum $K + V_{net}$ is called the total mechanical energy (E) of the system.

Principle of Conservation of Mechanical Energy:

The total mechanical energy of a system is conserved if the forces, doing work on it, are conservative.

This means that as the system changes configuration (position) and its potential energy changes, its kinetic energy changes by an equal and opposite amount, such that their sum remains constant throughout the motion.

$$ K_i + V_i = K_f + V_f $$

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Example: A ball of mass $m$ dropped from height $H$ (taking $V=0$ at ground level):

• At height $H$: $K_i = 0$ (starts from rest), $V_i = mgH$. Total energy $E = K_i + V_i = mgH$.
• At height $y$ ($0 < y < H$) with speed $v$: $K_y = \frac{1}{2}mv^2$, $V_y = mgy$. Total energy $E = K_y + V_y = \frac{1}{2}mv^2 + mgy$.
• At ground level ($y=0$) with speed $v_f$: $K_f = \frac{1}{2}mv_f^2$, $V_f = mg(0) = 0$. Total energy $E = K_f + V_f = \frac{1}{2}mv_f^2$.

By conservation of mechanical energy:

$$ mgH = \frac{1}{2}mv^2 + mgy = \frac{1}{2}mv_f^2 $$

From $mgH = \frac{1}{2}mv_f^2$, we get $v_f^2 = 2gH$, or $v_f = \sqrt{2gH}$ (final speed at ground). This matches kinematic results.

From $mgH = \frac{1}{2}mv^2 + mgy$, we get $\frac{1}{2}mv^2 = mg(H-y)$, or $v^2 = 2g(H-y)$ (speed at height $y$), also matching kinematic results. As potential energy decreases, kinetic energy increases, keeping the total mechanical energy constant.

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Answer:

The Potential Energy Of A Spring

The force exerted by an ideal spring is a classic example of a conservative force that varies with position. This force is described by Hooke's Law.

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For an ideal spring, the force it exerts ($\mathbf{F}_s$) is proportional to the displacement ($\mathbf{x}$) from its equilibrium (unstretched/uncompressed) position, and acts in the opposite direction to the displacement:

$$ \mathbf{F}_s = -k \mathbf{x} $$

In one dimension (along the x-axis), $F_s(x) = -kx$, where $k$ is the spring constant (units N/m), and $x$ is the displacement from equilibrium ($x=0$).

• If the spring is stretched ($x > 0$), $F_s$ is negative (pulling back towards equilibrium).
• If the spring is compressed ($x < 0$), $F_s$ is positive (pushing back towards equilibrium).
• At equilibrium ($x = 0$), $F_s = 0$.

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The work done by the spring force when the spring is stretched or compressed from an initial displacement $x_i$ to a final displacement $x_f$ is:

$$ W_s = \int_{x_i}^{x_f} F_s(x) \, dx = \int_{x_i}^{x_f} (-kx) \, dx = \left[-\frac{1}{2}kx^2\right]_{x_i}^{x_f} = -\frac{1}{2}kx_f^2 - (-\frac{1}{2}kx_i^2) $$ $$ W_s = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2 $$

This shows that the work done by the spring force depends only on the initial and final displacements, not on the path taken (consistent with a conservative force).

If the spring is stretched to $x_m$ and then returned to $x_m$, the work done by the spring force is $\frac{1}{2}kx_m^2 - \frac{1}{2}kx_m^2 = 0$. The work done in a cyclic process is zero.

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Since the spring force is conservative, we can associate a potential energy with it. The potential energy of a spring (V), choosing $V=0$ at the equilibrium position ($x=0$), is defined as:

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

We can verify that $-\frac{dV}{dx} = -\frac{d}{dx}(\frac{1}{2}kx^2) = -\frac{1}{2}k(2x) = -kx = F_s(x)$.

The work done by the spring force is also given by $W_s = V(x_i) - V(x_f) = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$, which matches the integrated result.

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For a mass $m$ attached to a spring and undergoing motion (like oscillations) without non-conservative forces, the total mechanical energy $E = K + V$ is conserved:

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

If the mass is released from rest at maximum extension $x_m$ (where $v=0$), the total energy is $E = \frac{1}{2}k x_m^2$. At any other position $x$ with speed $v$, the energy is $\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kx_m^2$. This shows the interconversion between kinetic and potential energy while the total remains constant.

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Answer:

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Answer:

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Remarks on Conservative Forces and Potential Energy:

• Time is not explicitly present in the potential energy function or the principle of energy conservation ($E=K+V$). It relates energy values at different positions.
• Not all forces are conservative (e.g., friction). When non-conservative forces do work, mechanical energy ($K+V$) is not conserved. The work done by non-conservative forces equals the change in mechanical energy: $W_{nc} = \Delta E$.
• The choice of the zero point of potential energy is arbitrary and convenient (e.g., ground for gravity, equilibrium for spring). Only *changes* in potential energy are physically significant.

Various Forms Of Energy : The Law Of Conservation Of Energy

Energy exists in many different forms besides mechanical energy (kinetic and potential).

Energy can be transformed from one form to another. The total energy of an isolated system remains constant, even if transformations occur.

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Some other forms of energy:

• Heat Energy: Often generated by friction or resistance. It is related to the internal energy of a substance, which is the sum of the kinetic and potential energies of its constituent molecules. When mechanical energy is "lost" due to non-conservative forces like friction, it is converted into heat.
• Chemical Energy: Stored in the bonds of molecules. Released or absorbed during chemical reactions (exothermic/endothermic reactions). It arises from the electrical forces between atoms and molecules.
• Electrical Energy: Energy associated with electric charges and currents.
• Light Energy: Energy carried by electromagnetic waves.
• Sound Energy: Energy transmitted through mechanical waves.
• Nuclear Energy: Energy stored in the nucleus of atoms, released during nuclear reactions (fission, fusion). It arises from the conversion of a small amount of mass into energy according to Einstein's mass-energy equivalence relation $E = mc^2$.

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Einstein's Mass-Energy Equivalence:

Einstein showed that mass and energy are interchangeable and are related by the equation:

$$ E = m c^2 $$

where $m$ is mass and $c$ is the speed of light in vacuum ($c \approx 3 \times 10^8$ m/s).

This implies that mass itself is a form of energy. In processes like nuclear reactions, a measurable amount of mass is converted into a large amount of energy (or vice versa). Even in chemical reactions, there's a tiny mass difference, but it's usually negligible compared to nuclear reactions.

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The Principle of Conservation of Energy:

This is one of the most fundamental principles in physics. It is a broader principle than the conservation of mechanical energy.

The total energy of an isolated system remains constant. Energy can be transformed from one form to another, but it cannot be created or destroyed.

An isolated system is one that does not exchange energy with its surroundings.

If we consider the universe as a whole, it can be viewed as an isolated system, and thus the total energy of the universe is constant.

This principle unifies various branches of physics and underlies many natural processes and technological applications involving energy transformations.

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Table 6.3 Approximate energy associated with various phenomena:

Description	Energy (Joules)
Motion of air molecule ($10^{-26}$ kg) at room temperature	$10^{-21}$
Typical binding energy of an electron in an atom	$10^{-18}$
Typical energy to break a chemical bond	$10^{-19}$ - $10^{-20}$
Molecular collision	$10^{-20}$
Energy released in fission of a U-235 atom	$200 \times 10^6 \text{ eV}$ ($\approx 3.2 \times 10^{-11}$ J)
Energy from burning 1 kg of coal	$\approx 3 \times 10^7$
Energy from the daily food intake of an average adult	$\approx 10^7$
Kinetic energy of a cricket ball	$\approx 10^2$
Kinetic energy of a car at 60 km/h	$\approx 3 \times 10^5$
Gravitational potential energy of water in a 1000 MW dam	$\approx 10^8$
Kinetic energy of a train at 60 km/h	$\approx 7 \times 10^7$
Energy released in the Hiroshima bomb	$\approx 10^{14}$
Annual electrical output of a large power plant	$\approx 10^{17}$
Energy released in a supernova explosion	$\approx 10^{44}$
Total energy radiated by the Sun in one second	$\approx 4 \times 10^{26}$
Total energy of the universe	Very large, but constant

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Answer:

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Common Misconception: Food energy values are often reported in 'calories' (with a capital C) in popular media, where 1 Calorie (with a capital C) means 1 kilocalorie (kcal). So, 2400 Calories in nutrition means 2400 kilocalories.

Power

Power (P) is defined as the rate at which work is done or the rate at which energy is transferred.

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Average Power ($P_{av}$) is the total work done ($W$) divided by the total time taken ($t$):

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

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Instantaneous Power (P) is the instantaneous rate of doing work. It is the limit of average power as the time interval approaches zero:

$$ P = \lim_{\Delta t \to 0} \frac{\Delta W}{\Delta t} = \frac{dW}{dt} $$

Since the work done over an infinitesimal displacement $d\mathbf{r}$ by a force $\mathbf{F}$ is $dW = \mathbf{F} \cdot d\mathbf{r}$, the instantaneous power can also be expressed as:

$$ P = \frac{\mathbf{F} \cdot d\mathbf{r}}{dt} = \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} $$ $$ P = \mathbf{F} \cdot \mathbf{v} $$

where $\mathbf{v}$ is the instantaneous velocity of the object on which the force $\mathbf{F}$ is acting.

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Power is a scalar quantity. Its dimensions are $[ML^2T^{-3}]$ (Work/Time). The SI unit of power is the watt (W), named after James Watt. $1 \text{ W} = 1 \text{ J/s}$.

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Other units of power:

• Horsepower (hp): $1 \text{ hp} = 746 \text{ W}$. Used for engines.
• Kilowatt (kW): $1 \text{ kW} = 1000 \text{ W}$.

Kilowatt-hour (kWh) is a unit of energy, not power. $1 \text{ kWh} = 1 \text{ kW} \times 1 \text{ hour} = 1000 \text{ J/s} \times 3600 \text{ s} = 3.6 \times 10^6$ J. This is the unit used in electricity bills.

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Answer:

Collisions

A collision is an event in which two or more bodies exert forces on each other for a relatively short period. Collisions are analyzed using conservation laws, particularly conservation of momentum and sometimes conservation of kinetic energy.

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Conservation of Momentum in Collisions:

During a collision, the forces between the colliding bodies are typically much larger than any external forces acting on the system. For the duration of the collision, the system of colliding bodies can be considered approximately isolated.

According to Newton's Third Law, the mutual forces between the colliding bodies ($\mathbf{F}_{12} = -\mathbf{F}_{21}$) form action-reaction pairs. By Newton's Second Law, the change in momentum of each body is due to the force exerted by the other body over the collision time $\Delta t$ ($\Delta \mathbf{p}_1 = \int \mathbf{F}_{12} dt$, $\Delta \mathbf{p}_2 = \int \mathbf{F}_{21} dt$).

Since $\mathbf{F}_{12} = -\mathbf{F}_{21}$, the total impulse on the system is $\int (\mathbf{F}_{12} + \mathbf{F}_{21}) dt = \int \mathbf{0} dt = \mathbf{0}$.

Therefore, the total change in momentum of the system is zero: $\Delta \mathbf{p}_{total} = \Delta \mathbf{p}_1 + \Delta \mathbf{p}_2 = \mathbf{0}$.

This means the total momentum of the system is conserved during the collision.

Total momentum before collision = Total momentum after collision

$$ \sum \mathbf{p}_{initial} = \sum \mathbf{p}_{final} $$

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Kinetic Energy in Collisions:

Unlike momentum, the total kinetic energy of the system is not always conserved during a collision. Some kinetic energy may be converted into other forms of energy (heat, sound, deformation energy).

Collisions are classified based on whether kinetic energy is conserved:

1. Elastic Collision: A collision in which the total kinetic energy of the system is conserved. This means the total kinetic energy *before* collision equals the total kinetic energy *after* collision. This typically happens when the deformation during impact is fully recovered, like in collisions between ideal elastic bodies.
2. Inelastic Collision: A collision in which the total kinetic energy of the system is not conserved. Some kinetic energy is lost (converted to other forms). Most real-world collisions are inelastic.
3. Completely Inelastic Collision: A specific type of inelastic collision where the colliding bodies stick together and move as a single unit after the collision. This is the maximum possible loss of kinetic energy in a collision consistent with momentum conservation.

Momentum is conserved in ALL types of collisions (elastic, inelastic, completely inelastic), assuming the system is isolated.

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Collisions in One Dimension (Head-on Collisions)

In a one-dimensional collision, the initial and final velocities of the particles are all along the same straight line.

Consider two particles $m_1$ and $m_2$ colliding head-on. Let their initial velocities be $v_{1i}$ and $v_{2i}$ (with signs indicating direction), and final velocities be $v_{1f}$ and $v_{2f}$.

Momentum Conservation:

$$ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} $$

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Kinetic Energy Conservation (for Elastic Collisions):

$$ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 $$

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For an elastic collision in 1D, with $m_2$ initially at rest ($v_{2i} = 0$), the conservation equations are:

1) $m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}$ 2) $\frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$

Solving these two equations simultaneously for the unknown final velocities $v_{1f}$ and $v_{2f}$ (in terms of $m_1, m_2, v_{1i}$):

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

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Special Cases (for elastic 1D collision with $m_2$ at rest):

• Equal Masses ($m_1 = m_2$): $$ v_{1f} = \left(\frac{m_1 - m_1}{m_1 + m_1}\right) v_{1i} = 0 $$ $$ v_{2f} = \left(\frac{2m_1}{m_1 + m_1}\right) v_{1i} = v_{1i} $$ The first mass stops, and the second mass moves off with the initial velocity of the first. (Momentum $m_1 v_{1i} + 0 = m_1(0) + m_1 v_{1i}$. Energy $\frac{1}{2} m_1 v_{1i}^2 + 0 = \frac{1}{2} m_1(0)^2 + \frac{1}{2} m_1 v_{1i}^2$).
• Target much heavier ($m_2 >> m_1$): $$ v_{1f} \approx \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_{1i} \approx \left(\frac{-m_2}{m_2}\right) v_{1i} = -v_{1i} $$ $$ v_{2f} \approx \left(\frac{2m_1}{m_1 + m_2}\right) v_{1i} \approx \left(\frac{2m_1}{m_2}\right) v_{1i} \approx 0 $$ The lighter mass bounces back with nearly its original speed, and the heavier mass remains almost at rest. (Think of a ball bouncing off a wall).
• Target much lighter ($m_1 >> m_2$): $$ v_{1f} \approx \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_{1i} \approx \left(\frac{m_1}{m_1}\right) v_{1i} = v_{1i} $$ $$ v_{2f} \approx \left(\frac{2m_1}{m_1 + m_2}\right) v_{1i} \approx \left(\frac{2m_1}{m_1}\right) v_{1i} = 2v_{1i} $$ The heavier mass continues with nearly its original velocity, and the lighter mass is knocked forward with approximately twice the initial velocity of the heavier mass. (Think of a car hitting a stationary bicycle).

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For a completely inelastic collision in 1D (with $m_2$ initially at rest, $v_{2i} = 0$), the bodies stick together, so they have a common final velocity $v_f$ ($v_{1f} = v_{2f} = v_f$).

Momentum conservation: $m_1 v_{1i} + m_2 (0) = (m_1 + m_2) v_f$.

$$ v_f = \frac{m_1 v_{1i}}{m_1 + m_2} $$

In this case, kinetic energy is not conserved. The loss in kinetic energy is $\Delta K = K_f - K_i = \frac{1}{2}(m_1+m_2)v_f^2 - \frac{1}{2}m_1 v_{1i}^2$. Substituting $v_f$ shows that $\Delta K$ is negative, indicating energy loss.

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Answer:

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Collisions in Two Dimensions

When colliding bodies move in a plane (or 3D space) before or after the collision, it's a two-dimensional (or three-dimensional) collision. Momentum conservation applies as a vector equation.

Consider a collision between mass $m_1$ (initial velocity $\mathbf{v}_{1i}$) and stationary mass $m_2$ (initial velocity $\mathbf{v}_{2i}=\mathbf{0}$). After the collision, they move with velocities $\mathbf{v}_{1f}$ and $\mathbf{v}_{2f}$ at angles $\theta_1$ and $\theta_2$ relative to the initial direction of $m_1$ (chosen as the x-axis).

Momentum Conservation (vector form):

$$ m_1 \mathbf{v}_{1i} = m_1 \mathbf{v}_{1f} + m_2 \mathbf{v}_{2f} $$

In component form (assuming the collision is in the x-y plane):

• x-component: $m_1 v_{1i} = m_1 v_{1f} \cos \theta_1 + m_2 v_{2f} \cos \theta_2$
• y-component: $0 = m_1 v_{1f} \sin \theta_1 + m_2 v_{2f} \sin \theta_2$

(Note: $\theta_1$ and $\theta_2$ might be defined such that one angle is positive and the other negative relative to the x-axis, reflecting opposite y-components of velocity after collision). If $\theta_1$ and $\theta_2$ are the magnitudes of the angles with the x-axis, and $v_{1fy}$ and $v_{2fy}$ have opposite signs, then the y-component equation would be $0 = m_1 v_{1f} \sin \theta_1 - m_2 v_{2f} \sin \theta_2$, as in the text's equation.

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If the collision is also elastic, total kinetic energy is conserved:

$$ \frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 $$

For a 2D elastic collision with $m_2$ initially at rest, we have 3 equations (2 from momentum, 1 from kinetic energy) and 4 unknowns ($v_{1f}, v_{2f}, \theta_1, \theta_2$). We need additional information (like one of the angles) to solve the problem completely.

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Scattering:

When particles interact via forces acting at a distance (like electromagnetic or gravitational forces), rather than direct physical contact, the event is often called scattering (e.g., a comet near the sun, an alpha particle near a nucleus). The final velocities and directions depend on the initial conditions and the nature of the interaction forces.

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