Chapter Notes

Introduction

In everyday language, we use terms like 'work', 'energy', and 'power' quite freely. For instance, studying for an exam is considered 'work'. In physics, these terms have very precise meanings.

• Work: In physics, work is only done when a force causes an object to move a certain distance. Pushing against a wall that doesn't move results in zero work, even though you might feel tired.
• Energy: This is the capacity to do work. A person with high stamina has a lot of energy, and in physics, energy is directly related to the ability to perform work.
• Power: This relates to how quickly work is done. A 'powerful' punch in boxing is one that is delivered very fast. This is close to the physics definition of power, which is the rate of doing work.

To understand these concepts mathematically, we first need to learn about the scalar product of vectors.

The Scalar Product

Many physical quantities like force and displacement are vectors, meaning they have both magnitude and direction. We can multiply vectors in two ways. One is the scalar product, and the other is the vector product (which will be covered in a later chapter).

The scalar product, also known as the dot product, of two vectors A and B gives a scalar (a number without direction) as the result. It is written as $\mathbf{A} \cdot \mathbf{B}$ and defined by the formula:

Scalar Product: $\mathbf{A} \cdot \mathbf{B} = AB \cos \theta$

Here, $A$ and $B$ are the magnitudes of the vectors, and $\theta$ is the angle between them.

The scalar product can be interpreted in two ways:

1. The magnitude of vector A multiplied by the component of vector B that is along the direction of A ($B \cos \theta$).
2. The magnitude of vector B multiplied by the component of vector A that is along the direction of B ($A \cos \theta$).

Properties of the Scalar Product:

• Commutative Law: The order of multiplication doesn't matter. $\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}$
• Distributive Law: It can be distributed over vector addition. $\mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = \mathbf{A} \cdot \mathbf{B} + \mathbf{A} \cdot \mathbf{C}$

Scalar Product with Unit Vectors: For the standard unit vectors $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$:

• The dot product of a unit vector with itself is 1 (since $\theta = 0^\circ$ and $\cos 0^\circ = 1$). $\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{k}} = 1$
• The dot product of two different unit vectors is 0 (since they are perpendicular, $\theta = 90^\circ$ and $\cos 90^\circ = 0$). $\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{k}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{i}} = 0$

Scalar Product in Component Form: If we have two vectors in component form: $\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}}$ $\mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}} + B_z \hat{\mathbf{k}}$

Their scalar product is: $\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$

Given

• Force vector, $\mathbf{F} = (3\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}})$ unit
• Displacement vector, $\mathbf{d} = (5\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 3\hat{\mathbf{k}})$ unit

To Find

• The angle $\theta$ between $\mathbf{F}$ and $\mathbf{d}$.
• The projection of $\mathbf{F}$ on $\mathbf{d}$.

Formula

$\mathbf{F} \cdot \mathbf{d} = F_x d_x + F_y d_y + F_z d_z$ $\mathbf{F} \cdot \mathbf{d} = Fd \cos \theta$ $F^2 = F_x^2 + F_y^2 + F_z^2$ $d^2 = d_x^2 + d_y^2 + d_z^2$ Projection of $\mathbf{F}$ on $\mathbf{d}$ is $F \cos \theta = \frac{\mathbf{F} \cdot \mathbf{d}}{d}$.

Solution

First, calculate the dot product $\mathbf{F} \cdot \mathbf{d}$: $\mathbf{F} \cdot \mathbf{d} = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16 \text{ unit}$

Next, find the magnitudes of $\mathbf{F}$ and $\mathbf{d}$: $F^2 = 3^2 + 4^2 + (-5)^2 = 9 + 16 + 25 = 50 \text{ unit}$ $F = \sqrt{50} \text{ unit}$ $d^2 = 5^2 + 4^2 + 3^2 = 25 + 16 + 9 = 50 \text{ unit}$ $d = \sqrt{50} \text{ unit}$

Now, find the angle $\theta$ using the dot product formula: $\cos \theta = \frac{\mathbf{F} \cdot \mathbf{d}}{Fd} = \frac{16}{\sqrt{50} \sqrt{50}} = \frac{16}{50} = 0.32$ $\theta = \cos^{-1}(0.32)$

The projection of $\mathbf{F}$ on $\mathbf{d}$ is the component of $\mathbf{F}$ along $\mathbf{d}$, which is $F \cos \theta$. $F \cos \theta = \frac{\mathbf{F} \cdot \mathbf{d}}{d} = \frac{16}{\sqrt{50}}$

Final Answer The angle between the force and displacement is $\theta = \cos^{-1}(0.32)$. The projection of $\mathbf{F}$ on $\mathbf{d}$ is $\frac{16}{\sqrt{50}}$ unit.

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

Let's start with a familiar equation of motion for an object moving in a straight line with constant acceleration: $v^2 - u^2 = 2as$ where $u$ is initial speed, $v$ is final speed, $a$ is acceleration, and $s$ is distance.

If we multiply both sides by $\frac{m}{2}$ (where $m$ is the mass of the object), we get: $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mas$ From Newton's Second Law, we know that force $F = ma$. So, we can replace $mas$ with $Fs$: $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = Fs$

This equation gives us the foundation for defining two important concepts:

1. Kinetic Energy (K): The quantity $\frac{1}{2}mv^2$ is called the kinetic energy of the object. It's the energy an object possesses due to its motion.
2. Work (W): The quantity on the right side, $Fs$, is the product of the force and the displacement. This is defined as the work done by the force on the object. In vector form, this is $W = \mathbf{F} \cdot \mathbf{d}$.

With these definitions, the equation becomes: $K_f - K_i = W$ where $K_i$ is the initial kinetic energy and $K_f$ is the final kinetic energy.

This relationship is known as the Work-Energy Theorem. It states that the change in the kinetic energy of an object is equal to the net work done on it by all forces.

Given

• Mass of raindrop, $m = 1.00 \text{ g} = 10^{-3} \text{ kg}$
• Height, $h = 1.00 \text{ km} = 10^3 \text{ m}$
• Initial speed, $u = 0$ (assumed at rest)
• Final speed, $v = 50.0 \text{ m s}^{-1}$
• Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$

To Find

(a) Work done by gravity, $W_g$ (b) Work done by resistive force, $W_r$

Formula

• Change in kinetic energy: $\Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$
• Work done by gravity: $W_g = mgh$
• Work-Energy Theorem: $\Delta K = W_g + W_r$

Solution

(a) Calculate the work done by gravity

The change in kinetic energy of the drop is: $\Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2} \times 10^{-3} \text{ kg} \times (50 \text{ m s}^{-1})^2$ $\Delta K = \frac{1}{2} \times 10^{-3} \times 2500 = 1.25 \text{ J}$

The work done by the gravitational force is: $W_g = mgh = (10^{-3} \text{ kg}) \times (10 \text{ m s}^{-2}) \times (10^3 \text{ m})$ $W_g = 10.0 \text{ J}$

Answer for part (a) = $10.0$ J

(b) Calculate the work done by the resistive force

According to the work-energy theorem, the total work done by all forces (gravity and resistance) equals the change in kinetic energy. $\Delta K = W_g + W_r$ We can rearrange this to find $W_r$: $W_r = \Delta K - W_g$ $W_r = 1.25 \text{ J} - 10.0 \text{ J} = -8.75 \text{ J}$

The work done by the resistive force is negative because it acts in the opposite direction to the raindrop's motion.

Answer for part (b) = $-8.75$ J

Work

Work is done when a force causes a displacement. For a constant force $\mathbf{F}$ acting on an object that undergoes a displacement $\mathbf{d}$, the work done is defined as the product of the magnitude of the displacement and the component of the force in the direction of the displacement.

Mathematically, this is expressed using the scalar product: $W = (F \cos \theta)d = \mathbf{F} \cdot \mathbf{d}$ Here, $\theta$ is the angle between the force vector and the displacement vector.

Conditions for Zero Work: No work is done if:

1. Displacement is zero ($d=0$): A weightlifter holding a heavy weight steady on his shoulders does no work on the weight, no matter how much force he applies.
2. Force is zero ($F=0$): An object sliding on a frictionless horizontal surface may move a large distance, but with no horizontal force acting on it, no work is done.
3. Force and displacement are perpendicular ($\theta = 90^\circ$): Since $\cos 90^\circ = 0$, the work done is zero. For example, the Earth's gravitational force does no work on the Moon in a perfectly circular orbit, because the force is always perpendicular to the Moon's instantaneous displacement.

Positive and Negative Work:

• Positive Work: If the angle $\theta$ is between $0^\circ$ and $90^\circ$, $\cos\theta$ is positive, and the work done is positive. This means the force helps the motion.
• Negative Work: If the angle $\theta$ is between $90^\circ$ and $180^\circ$, $\cos\theta$ is negative, and the work done is negative. This means the force opposes the motion. Frictional force always does negative work.

The SI unit for work is the joule (J). The dimensions of work are $[\text{ML}^2\text{T}^{-2}]$.

Given

• Displacement, $d = 10$ m
• Stopping force, $F = 200$ N
• Angle between force and displacement, $\theta = 180^\circ$ (since the force opposes motion)

To Find

(a) Work done by the road on the cycle, $W_r$ (b) Work done by the cycle on the road, $W_c$

Formula

$W = Fd \cos \theta$

Solution

(a) Work done by the road on the cycle

The force is from the road, and the displacement is that of the cycle. $W_r = (200 \text{ N}) \times (10 \text{ m}) \times \cos(180^\circ)$ Since $\cos(180^\circ) = -1$: $W_r = 200 \times 10 \times (-1) = -2000 \text{ J}$ This negative work is what brings the cycle to a stop.

Answer for part (a) = $-2000$ J

(b) Work done by the cycle on the road

According to Newton's Third Law, the cycle exerts an equal and opposite force of 200 N on the road. However, the road does not move. Its displacement is zero. $W_c = (200 \text{ N}) \times (0 \text{ m}) \times \cos \theta = 0 \text{ J}$

Answer for part (b) = $0$ J

Kinetic Energy

As defined earlier, the kinetic energy of an object of mass $m$ moving with velocity $\mathbf{v}$ is: $K = \frac{1}{2} m \mathbf{v} \cdot \mathbf{v} = \frac{1}{2} m v^2$ Kinetic energy is a scalar quantity and is a measure of the work an object can do because of its motion. For example, the kinetic energy of flowing water is used to generate electricity, and the kinetic energy of wind is used by sailing ships.

Given

• Mass of bullet, $m = 50.0 \text{ g} = 0.05 \text{ kg}$
• Initial speed, $v_i = 200 \text{ m s}^{-1}$
• Final kinetic energy, $K_f = 10\%$ of initial kinetic energy, $K_i$

To Find

• Emergent speed of the bullet, $v_f$

Formula

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

Solution

First, calculate the initial kinetic energy ($K_i$): $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times (0.05 \text{ kg}) \times (200 \text{ m s}^{-1})^2$ $K_i = \frac{1}{2} \times 0.05 \times 40000 = 1000 \text{ J}$

The final kinetic energy ($K_f$) is 10% of this value: $K_f = 0.10 \times K_i = 0.10 \times 1000 \text{ J} = 100 \text{ J}$

Now, use the kinetic energy formula to find the final speed ($v_f$): $K_f = \frac{1}{2} m v_f^2$ $100 \text{ J} = \frac{1}{2} \times (0.05 \text{ kg}) \times v_f^2$ $v_f^2 = \frac{2 \times 100 \text{ J}}{0.05 \text{ kg}} = \frac{200}{0.05} = 4000$ $v_f = \sqrt{4000} \approx 63.2 \text{ m s}^{-1}$

Final Answer The emergent speed of the bullet is approximately $63.2 \text{ m s}^{-1}$.

Work Done by a Variable Force

In most real-world situations, forces are not constant. For example, the force of a spring changes as it is stretched. To calculate the work done by a variable force, we can use a graphical method or calculus.

Imagine a force $F(x)$ that changes with position $x$. We can divide the total displacement into tiny segments, $\Delta x$. Over each tiny segment, the force is almost constant. The small amount of work done, $\Delta W$, is: $\Delta W = F(x) \Delta x$ This corresponds to the area of a small rectangle on a Force vs. Displacement graph.

To find the total work done from an initial position $x_i$ to a final position $x_f$, we sum the areas of all these small rectangles: $W \cong \sum_{x_i}^{x_f} F(x) \Delta x$ In the limit as $\Delta x$ approaches zero, this sum becomes a definite integral, which gives the exact area under the curve.

Work done by a variable force: $W = \int_{x_i}^{x_f} F(x) dx$

Given

• Applied force $F$: 100 N for $0 \le x \le 10$ m.
• Applied force $F$: Reduces linearly from 100 N to 50 N for $10 < x \le 20$ m.
• Frictional force $f$: -50 N (constant, opposes motion).
• Total displacement: 20 m.

To Find

• Work done by the woman, $W_F$.
• Work done by the frictional force, $W_f$.

Formula

Work is the area under the Force-Displacement graph.

• Area of rectangle = length $\times$ width
• Area of trapezium = $\frac{1}{2} \times (\text{sum of parallel sides}) \times$ height

Solution

The work done is the area under the F-x graph.

Work done by the woman ($W_F$) The area can be split into a rectangle (from 0 to 10 m) and a trapezium (from 10 to 20 m).

• Area of rectangle (ABCD) = $100 \text{ N} \times 10 \text{ m} = 1000 \text{ J}$
• Area of trapezium (CEID) = $\frac{1}{2} (100 \text{ N} + 50 \text{ N}) \times (20-10) \text{ m} = \frac{1}{2} \times 150 \times 10 = 750 \text{ J}$

Total work done by the woman: $W_F = 1000 \text{ J} + 750 \text{ J} = 1750 \text{ J}$

Work done by the frictional force ($W_f$) The frictional force is constant at -50 N over the entire 20 m displacement. The area is a rectangle below the x-axis.

• Area of rectangle (AGHI) = $(-50 \text{ N}) \times 20 \text{ m} = -1000 \text{ J}$

Final Answer The work done by the woman is $1750$ J. The work done by the frictional force is $-1000$ J.

The Work-Energy Theorem for a Variable Force

The work-energy theorem ($K_f - K_i = W$) also holds true for variable forces. We can prove this using calculus.

The rate of change of kinetic energy with time is: $\frac{dK}{dt} = \frac{d}{dt} \left(\frac{1}{2}mv^2\right)$ Using the chain rule, this becomes: $\frac{dK}{dt} = m \frac{dv}{dt} v$ We know that acceleration $a = \frac{dv}{dt}$ and from Newton's Second Law, $F = ma$. So, $m \frac{dv}{dt} = F$. $\frac{dK}{dt} = Fv$ Since velocity $v = \frac{dx}{dt}$, we can write: $\frac{dK}{dt} = F \frac{dx}{dt}$ Multiplying by $dt$, we get: $dK = F dx$ To find the total change in kinetic energy over a displacement from $x_i$ to $x_f$, we integrate both sides: $\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F dx$ The left side is $K_f - K_i$, and the right side is the definition of work, $W$. $K_f - K_i = W$ This proves the work-energy theorem for a variable force in one dimension.

Given

• Mass, $m = 1$ kg
• Initial speed, $v_i = 2 \text{ m s}^{-1}$
• Retarding force, $F_r = \frac{-k}{x}$ for $0.1 < x < 2.01$ m
• Constant, $k = 0.5$ J

To Find

• Final kinetic energy, $K_f$
• Final speed, $v_f$

Formula

• Work-Energy Theorem: $K_f - K_i = W = \int_{x_i}^{x_f} F(x) dx$
• Initial kinetic energy: $K_i = \frac{1}{2}mv_i^2$

Solution

First, calculate the initial kinetic energy: $K_i = \frac{1}{2}(1 \text{ kg})(2 \text{ m s}^{-1})^2 = 2 \text{ J}$

Now, use the work-energy theorem. The work done by the retarding force is: $W = \int_{0.1}^{2.01} \frac{-k}{x} dx = -k [\ln(x)]_{0.1}^{2.01}$ $W = -k (\ln(2.01) - \ln(0.1)) = -k \ln\left(\frac{2.01}{0.1}\right) = -0.5 \ln(20.1)$ Given $\ln(20.1) \approx 3.00$, the work done is $W \approx -0.5 \times 3.00 = -1.5 \text{ J}$. (Using the value from the source, $0.5 \ln(20.1) = 1.5$) So, $W = -1.5 \text{ J}$.

Now apply the work-energy theorem to find $K_f$: $K_f - K_i = W$ $K_f = K_i + W = 2 \text{ J} - 1.5 \text{ J} = 0.5 \text{ J}$

Finally, calculate the final speed $v_f$: $K_f = \frac{1}{2}mv_f^2$ $0.5 \text{ J} = \frac{1}{2}(1 \text{ kg})v_f^2$ $v_f^2 = 1 \implies v_f = 1 \text{ m s}^{-1}$

Final Answer The final kinetic energy is $0.5$ J and the final speed is $1 \text{ m s}^{-1}$.

The Concept of Potential Energy

Potential energy is 'stored' energy that an object has due to its position or configuration. For example, a stretched bowstring has potential energy, which is converted into the kinetic energy of the arrow when released.

This concept applies only to a special class of forces called conservative forces. For these forces, the work done against them gets stored as potential energy.

Gravitational Potential Energy: Consider lifting a ball of mass $m$ to a height $h$. The work you do against the gravitational force ($mg$) is $mgh$. This work is stored as gravitational potential energy, denoted by $V(h)$. $V(h) = mgh$ If you release the ball, this stored potential energy is converted back into kinetic energy as it falls.

Conservative Forces and Potential Energy: A force $F(x)$ is conservative if it can be expressed as the negative derivative of a potential energy function $V(x)$: $F(x) = -\frac{dV}{dx}$ This relationship implies that the work done by a conservative force only depends on the initial and final positions, not on the path taken between them. $W = \int_{x_i}^{x_f} F(x) dx = -\int_{V_i}^{V_f} dV = V_i - V_f$ Forces like gravity and the spring force are conservative. Friction is a non-conservative force because the work done by friction depends on the length of the path taken.

The Conservation of Mechanical Energy

For a system where only conservative forces are doing work, the total mechanical energy is conserved (remains constant).

The total mechanical energy (E) is the sum of the kinetic energy (K) and the potential energy (V) of the system. $E = K + V$ Let's see why it's conserved. From the work-energy theorem, the change in kinetic energy is: $\Delta K = W$ For a conservative force, the work done is equal to the negative change in potential energy: $W = -\Delta V$ Combining these, we get: $\Delta K = -\Delta V \implies \Delta K + \Delta V = 0$ This can be written as: $\Delta(K+V) = 0$ This equation means that the change in the total mechanical energy $(K+V)$ is zero. Therefore, the total mechanical energy is constant.

Principle of Conservation of Mechanical Energy: $K_i + V_i = K_f + V_f$ The total mechanical energy of a system remains constant if the forces doing work on it are conservative.

Example: A Freely Falling Body Consider a ball of mass $m$ dropped from a height $H$. Let's define the potential energy to be zero at the ground level ($V=0$ at $h=0$).

• At height H (top): The ball is at rest, so $K=0$. The potential energy is $V = mgH$. Total Energy $E_H = 0 + mgH = mgH$.
• At height h (mid-fall): The potential energy is $V = mgh$. The kinetic energy is $K = \frac{1}{2}mv_h^2$. Total Energy $E_h = \frac{1}{2}mv_h^2 + mgh$.
• At height 0 (ground): The potential energy is $V=0$. The kinetic energy is $K = \frac{1}{2}mv_f^2$. Total Energy $E_0 = \frac{1}{2}mv_f^2 + 0 = \frac{1}{2}mv_f^2$.

Since gravity is a conservative force, mechanical energy is conserved: $E_H = E_h = E_0$. $mgH = \frac{1}{2}mv_h^2 + mgh = \frac{1}{2}mv_f^2$ From this, we can find the speed at any point. For example, equating the energy at the top and the bottom gives $mgH = \frac{1}{2}mv_f^2$, which simplifies to $v_f = \sqrt{2gH}$.

Given

• Mass of bob = $m$
• Length of string = $L$
• Potential energy at lowest point A is zero.
• Tension at highest point C is zero ($T_C=0$).

To Find

(i) Initial velocity, $v_o$ (ii) Speeds at B and C, $v_B$ and $v_C$ (iii) Ratio of kinetic energies, $K_B / K_C$

Formula

• Conservation of Mechanical Energy: $E_A = E_B = E_C$
• Newton's Second Law for circular motion: $F_{net} = \frac{mv^2}{L}$

Solution

The only force doing work is gravity (conservative), as the tension is always perpendicular to the displacement. So, mechanical energy is conserved. Let's set the potential energy $V=0$ at point A.

(i) Find the initial velocity $v_o$

• At point A (lowest): $V_A = 0$, $K_A = \frac{1}{2}mv_o^2$. Total energy $E = \frac{1}{2}mv_o^2$.
• At point C (highest): The height is $2L$, so $V_C = mg(2L)$. The kinetic energy is $K_C = \frac{1}{2}mv_C^2$. Total energy $E = \frac{1}{2}mv_C^2 + 2mgL$. At point C, the string goes slack ($T_C=0$). The only force providing the centripetal acceleration is gravity. $mg = \frac{mv_C^2}{L} \implies v_C^2 = gL$ Substitute this into the energy equation for C: $E = \frac{1}{2}m(gL) + 2mgL = \frac{5}{2}mgL$ By conservation of energy, $E_A = E_C$: $\frac{1}{2}mv_o^2 = \frac{5}{2}mgL \implies v_o^2 = 5gL$ $v_o = \sqrt{5gL}$

Answer for part (i) = $\sqrt{5gL}$

(ii) Find the speeds at B and C

From the calculation above, the speed at C is: $v_C = \sqrt{gL}$

• At point B (horizontal): The height is $L$, so $V_B = mgL$. The kinetic energy is $K_B = \frac{1}{2}mv_B^2$. Total energy $E_B = \frac{1}{2}mv_B^2 + mgL$. By conservation of energy, $E_A = E_B$: $\frac{1}{2}mv_o^2 = \frac{1}{2}mv_B^2 + mgL$ Substitute $v_o^2 = 5gL$: $\frac{1}{2}m(5gL) = \frac{1}{2}mv_B^2 + mgL$ $\frac{5}{2}mgL - mgL = \frac{1}{2}mv_B^2 \implies \frac{3}{2}mgL = \frac{1}{2}mv_B^2$ $v_B^2 = 3gL \implies v_B = \sqrt{3gL}$

Answer for part (ii) = $v_B = \sqrt{3gL}$ and $v_C = \sqrt{gL}$

(iii) Find the ratio of kinetic energies

$K_B = \frac{1}{2}mv_B^2 = \frac{1}{2}m(3gL) = \frac{3}{2}mgL$ $K_C = \frac{1}{2}mv_C^2 = \frac{1}{2}m(gL) = \frac{1}{2}mgL$ The ratio is: $\frac{K_B}{K_C} = \frac{\frac{3}{2}mgL}{\frac{1}{2}mgL} = 3$

Answer for part (iii) = $3:1$

The Potential Energy of a Spring

The force exerted by an ideal spring is a variable, conservative force. It follows Hooke's Law: $F_s = -kx$ where $k$ is the spring constant (a measure of the spring's stiffness, in $\text{N m}^{-1}$), and $x$ is the displacement from the equilibrium position. The negative sign indicates that the spring force always acts to restore the spring to its equilibrium position.

Since the spring force is conservative, we can define a potential energy for it. The work done by the spring force when it is stretched from 0 to a displacement $x_m$ is: $W_s = \int_0^{x_m} F_s dx = \int_0^{x_m} -kx dx = -\frac{1}{2}kx_m^2$ The potential energy stored in the spring is the negative of the work done by the spring force. We define the potential energy to be zero at the equilibrium position ($x=0$).

Elastic Potential Energy: $V(x) = \frac{1}{2}kx^2$ For a mass attached to a spring, the total mechanical energy is conserved (if there is no friction): $E = K + V = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}$ When the spring is at its maximum stretch ($x=x_m$), the velocity is zero, so all the energy is potential: $E = \frac{1}{2}kx_m^2$. When the mass passes through the equilibrium position ($x=0$), the potential energy is zero, and the speed is maximum ($v=v_m$). All the energy is kinetic: $E = \frac{1}{2}mv_m^2$. Therefore, $\frac{1}{2}mv_m^2 = \frac{1}{2}kx_m^2$.

Given

• Mass of car, $m = 1000$ kg
• Speed of car, $v = 18.0 \text{ km/h} = 5 \text{ m s}^{-1}$
• Spring constant, $k = 6.25 \times 10^3 \text{ N m}^{-1}$

To Find

• Maximum compression of the spring, $x_m$

Formula

• Conservation of Mechanical Energy: $K_i + V_i = K_f + V_f$
• Kinetic Energy: $K = \frac{1}{2}mv^2$
• Spring Potential Energy: $V = \frac{1}{2}kx^2$

Solution

At the moment of maximum compression, the car comes to a momentary stop ($v_f = 0$), so its kinetic energy is zero. All of its initial kinetic energy has been converted into the potential energy of the spring.

Initial state: Car is moving, spring is uncompressed. $K_i = \frac{1}{2}mv^2$, $V_i = 0$

Final state: Car is stopped, spring is maximally compressed. $K_f = 0$, $V_f = \frac{1}{2}kx_m^2$

By conservation of energy: $K_i = V_f$ $\frac{1}{2}mv^2 = \frac{1}{2}kx_m^2$ Calculate the initial kinetic energy: $K_i = \frac{1}{2} \times (1000 \text{ kg}) \times (5 \text{ m s}^{-1})^2 = \frac{1}{2} \times 1000 \times 25 = 12500 \text{ J} = 1.25 \times 10^4 \text{ J}$ Now set this equal to the final potential energy: $1.25 \times 10^4 \text{ J} = \frac{1}{2} (6.25 \times 10^3 \text{ N m}^{-1}) x_m^2$ $x_m^2 = \frac{2 \times 1.25 \times 10^4}{6.25 \times 10^3} = \frac{2.5 \times 10^4}{0.625 \times 10^4} = 4$ $x_m = \sqrt{4} = 2.00 \text{ m}$

Final Answer The maximum compression of the spring is $2.00$ m.

Presence of Non-Conservative Forces

If a non-conservative force like friction is present, mechanical energy is not conserved. The work done by the non-conservative force ($W_{nc}$) equals the change in the total mechanical energy of the system. $\Delta E = E_f - E_i = W_{nc}$

Given

• Mass of car, $m = 1000$ kg
• Initial speed, $v = 5 \text{ m s}^{-1}$
• Spring constant, $k = 6.25 \times 10^3 \text{ N m}^{-1}$
• Coefficient of friction, $\mu = 0.5$

To Find

• Maximum compression, $x_m$

Formula

• Work-Energy Theorem: $\Delta K = W_{net}$
• Work done by spring: $W_s = -\frac{1}{2}kx_m^2$
• Work done by friction: $W_f = -f \times x_m = -\mu mg x_m$

Solution

The change in kinetic energy is from its initial value to zero. $\Delta K = K_f - K_i = 0 - \frac{1}{2}mv^2 = -1.25 \times 10^4 \text{ J}$ The net work done is the sum of the work done by the spring force and the frictional force. $W_{net} = W_s + W_f = -\frac{1}{2}kx_m^2 - \mu mg x_m$ According to the work-energy theorem, $\Delta K = W_{net}$: $-\frac{1}{2}mv^2 = -\frac{1}{2}kx_m^2 - \mu mg x_m$ Multiplying by -1 gives: $\frac{1}{2}mv^2 = \frac{1}{2}kx_m^2 + \mu mg x_m$ Substitute the values: $1.25 \times 10^4 = \frac{1}{2}(6.25 \times 10^3)x_m^2 + (0.5)(1000)(10)x_m$ $12500 = 3125 x_m^2 + 5000 x_m$ Divide by 125: $100 = 25 x_m^2 + 40 x_m$ Rearrange into a quadratic equation: $25x_m^2 + 40x_m - 100 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x_m = \frac{-40 + \sqrt{40^2 - 4(25)(-100)}}{2(25)}$ $x_m = \frac{-40 + \sqrt{1600 + 10000}}{50} = \frac{-40 + \sqrt{11600}}{50} \approx \frac{-40 + 107.7}{50} = \frac{67.7}{50} \approx 1.35 \text{ m}$ (We take the positive root since compression must be a positive distance).

Final Answer The maximum compression of the spring is $1.35$ m.

Power

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

• Average Power ($P_{av}$) is the total work done divided by the total time taken. $P_{av} = \frac{W}{t}$
• Instantaneous Power (P) is the rate of doing work at a specific moment. $P = \frac{dW}{dt}$ Since work $dW = \mathbf{F} \cdot d\mathbf{r}$, we can also express instantaneous power as: $P = \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = \mathbf{F} \cdot \mathbf{v}$ where $\mathbf{v}$ is the instantaneous velocity.

Power is a scalar quantity. Its SI unit is the watt (W), where $1 \text{ W} = 1 \text{ J s}^{-1}$. Another common unit is horsepower (hp), where $1 \text{ hp} = 746 \text{ W}$.

Given

• Total mass, $m = 1800$ kg
• Constant speed, $v = 2 \text{ m s}^{-1}$
• Frictional force, $F_f = 4000$ N
• Acceleration due to gravity, $g \approx 10 \text{ m s}^{-2}$

To Find

• Minimum power delivered by the motor, $P$.

Formula

$P = \mathbf{F} \cdot \mathbf{v}$

Solution

To move the elevator up at a constant speed, the motor must provide an upward force $F$ that exactly balances the downward forces. The downward forces are gravity ($mg$) and friction ($F_f$). $F_{down} = mg + F_f = (1800 \text{ kg} \times 10 \text{ m s}^{-2}) + 4000 \text{ N}$ $F_{down} = 18000 \text{ N} + 4000 \text{ N} = 22000 \text{ N}$ The motor must supply an upward force $F = 22000$ N. Since the force and velocity are in the same direction, the power delivered is: $P = F \times v = 22000 \text{ N} \times 2 \text{ m s}^{-1} = 44000 \text{ W}$ To convert this to horsepower: $P_{hp} = \frac{44000 \text{ W}}{746 \text{ W/hp}} \approx 59 \text{ hp}$

Final Answer The minimum power delivered by the motor is $44000$ W or approximately $59$ hp.

Collisions

A collision is an event where two or more bodies exert forces on each other for a relatively short time. In any collision, one fundamental principle always holds true: the conservation of total linear momentum.

Conservation of Linear Momentum in Collisions: When two objects collide, they exert equal and opposite forces on each other (Newton's Third Law, $\mathbf{F}_{12} = -\mathbf{F}_{21}$). The change in momentum (impulse) for each object is: $\Delta \mathbf{p}_1 = \mathbf{F}_{12} \Delta t$ $\Delta \mathbf{p}_2 = \mathbf{F}_{21} \Delta t$ Since the forces are equal and opposite, the changes in momentum are also equal and opposite: $\Delta \mathbf{p}_1 = -\Delta \mathbf{p}_2 \implies \Delta \mathbf{p}_1 + \Delta \mathbf{p}_2 = 0$ This means the total change in momentum of the system is zero, so the total momentum before the collision is equal to the total momentum after the collision.

Elastic and Inelastic Collisions

While momentum is always conserved, kinetic energy may not be.

• Elastic Collision: A collision where the total kinetic energy of the system is also conserved. The objects bounce off each other without any loss of kinetic energy (though it might be temporarily converted to potential energy during the impact).
• Inelastic Collision: A collision where some of the kinetic energy is lost, usually converted into other forms like heat, sound, or deformation of the objects.
• Completely Inelastic Collision: An inelastic collision where the objects stick together and move with a common final velocity after the impact. This is where the maximum possible kinetic energy is lost.

Collisions in One Dimension

Let's consider a mass $m_1$ with initial velocity $v_{1i}$ hitting a stationary mass $m_2$ ($v_{2i}=0$) in a head-on collision.

Completely Inelastic Collision: The two masses stick together and move with a final velocity $v_f$.

• Momentum Conservation: $m_1 v_{1i} = (m_1 + m_2) v_f$ $v_f = \frac{m_1}{m_1 + m_2} v_{1i}$
• Loss in Kinetic Energy: $\Delta K = K_i - K_f = \frac{1}{2}m_1 v_{1i}^2 - \frac{1}{2}(m_1+m_2)v_f^2$ $\Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} v_{1i}^2$ Since masses and velocity squared are positive, $\Delta K$ is always positive, meaning kinetic energy is always lost.

Elastic Collision: Both momentum and kinetic energy are conserved.

1. Momentum Conservation: $m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}$
2. Kinetic Energy Conservation: $\frac{1}{2}m_1 v_{1i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$

Solving these two equations gives the final velocities: $v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i}$ $v_{2f} = \frac{2m_1}{m_1 + m_2} v_{1i}$

Special Cases for Elastic Collisions:

• Case I: Equal Masses ($m_1 = m_2$): $v_{1f} = 0$ $v_{2f} = v_{1i}$ The first mass stops completely, and the second mass moves off with the initial velocity of the first. This is often seen in billiards.
• Case II: A heavy target ($m_2 \gg m_1$): $v_{1f} \approx -v_{1i}$ $v_{2f} \approx 0$ The light object bounces back with nearly its original speed, and the heavy object barely moves. Think of a ball bouncing off a wall.

Given

• Elastic collision between a neutron ($m_1$) and a moderator nucleus ($m_2$).

To Find

• The fractional kinetic energy lost by the neutron.

Formula

• Initial KE of neutron: $K_{1i} = \frac{1}{2}m_1 v_{1i}^2$
• Final KE of neutron: $K_{1f} = \frac{1}{2}m_1 v_{1f}^2$
• Final velocity of neutron: $v_{1f} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v_{1i}$

Solution

The fractional kinetic energy remaining for the neutron is $f_1 = K_{1f} / K_{1i}$. $f_1 = \frac{\frac{1}{2}m_1 v_{1f}^2}{\frac{1}{2}m_1 v_{1i}^2} = \frac{v_{1f}^2}{v_{1i}^2}$ Substitute the expression for $v_{1f}$: $f_1 = \left(\frac{\frac{m_1 - m_2}{m_1 + m_2} v_{1i}}{v_{1i}}\right)^2 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2$ The fractional energy transferred to the moderator nucleus is $f_2 = 1 - f_1$. $f_2 = 1 - \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 = \frac{(m_1+m_2)^2 - (m_1-m_2)^2}{(m_1+m_2)^2} = \frac{4m_1 m_2}{(m_1 + m_2)^2}$ Let's analyze for deuterium, where $m_2 \approx 2m_1$. $f_1 = \left(\frac{m_1 - 2m_1}{m_1 + 2m_1}\right)^2 = \left(\frac{-m_1}{3m_1}\right)^2 = \frac{1}{9}$ The fraction of energy transferred is $f_2 = 1 - f_1 = 1 - 1/9 = 8/9$.

Final Answer When a neutron collides with a deuterium nucleus, it transfers $8/9$ (almost 90%) of its kinetic energy, effectively slowing it down. This is why light nuclei are used as moderators.

Collisions in Two Dimensions

When objects collide at an angle (not head-on), the collision is two-dimensional. We must apply the conservation of momentum in both the x and y directions.

Consider a mass $m_1$ with initial velocity $v_{1i}$ along the x-axis hitting a stationary mass $m_2$. After the collision, $m_1$ moves off at an angle $\theta_1$ with velocity $v_{1f}$, and $m_2$ moves off at an angle $\theta_2$ with velocity $v_{2f}$.

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

We have two equations but four unknowns ($v_{1f}, v_{2f}, \theta_1, \theta_2$). If the collision is elastic, we have a third equation:

• Kinetic Energy Conservation: $\frac{1}{2}m_1 v_{1i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$

Even with three equations, we still have four unknowns. To solve the problem, we must know at least one of the final parameters, such as one of the scattering angles.

Given

• Equal masses, $m_1 = m_2 = m$
• Elastic collision
• Initial velocity of target, $v_{2i} = 0$
• Final angle of target, $\theta_2 = 37^\circ$

To Find

• Final angle of cue ball, $\theta_1$

Formula

• Vector momentum conservation: $\mathbf{v}_{1i} = \mathbf{v}_{1f} + \mathbf{v}_{2f}$
• Kinetic energy conservation: $v_{1i}^2 = v_{1f}^2 + v_{2f}^2$

Solution

From vector momentum conservation, we can square the equation using the dot product: $\mathbf{v}_{1i} \cdot \mathbf{v}_{1i} = (\mathbf{v}_{1f} + \mathbf{v}_{2f}) \cdot (\mathbf{v}_{1f} + \mathbf{v}_{2f})$ $v_{1i}^2 = \mathbf{v}_{1f} \cdot \mathbf{v}_{1f} + \mathbf{v}_{2f} \cdot \mathbf{v}_{2f} + 2 \mathbf{v}_{1f} \cdot \mathbf{v}_{2f}$ $v_{1i}^2 = v_{1f}^2 + v_{2f}^2 + 2 v_{1f} v_{2f} \cos(\theta_1 + \theta_2)$ The angle between the final velocity vectors is $\theta_1 + \theta_2$.