Work, Energy, & Power

Fundamentals of Work

Work describes the movement of an object by an external force over a distance. During this process, displacement occurs in the direction of motion, resulting in a transfer of energy. To honor James Prescott Joule for his contribution to the work and energy change relationship, the standard unit of energy was named the Joule ($\text{J}$).

The mathematical formula for work is:

$W = F\,d$

Where:

• $F$ is the force exerted on the object

• $d$ is the distance moved

• $W$ is the work done

When a force is exerted at an angle perpendicular to the direction of motion/displacement, the formula includes the angle component:

$W = F\,d\cos(\Theta)$

Work is also performed through actions such as:

• Compression of a gas

• Spinning of a shaft

• Particle motion caused by an external force

Concept and Forms of Energy

Energy is defined as the capacity to do work, which represents the ability of an object to produce a change in itself as well as in the world around it.

Energy exists in various forms:

• Potential energy

• Kinetic energy

• Thermal energy

• Chemical energy

• Nuclear energy

• Mechanical energy

• Sound energy

• Light energy

Every form of energy can change form into any other form.

The law of conservation of energy (the first law of thermodynamics) states that energy cannot be created or destroyed. As long as no external forces act upon a system, energy can only change form; therefore, the total amount of energy remains constant and is conserved within the system.

Primary energy forms and behaviors include:

• Potential Energy: Stored energy. For example, a marble situated at the top of an inclined plane possesses stored potential energy to roll down the incline. When the marble is pushed down the inclined plane, that stored energy transforms into kinetic energy.

• Kinetic Energy: The energy of motion. An example is the motion of the marble traveling down the inclined plane until it comes to a stop as a result of opposing frictional forces. This motion is converted into thermal energy, heating both the marble and the incline.

• Thermal Energy: The internal energy of movement of atoms that causes objects to experience an increase in temperature.

The Work-Energy Theorem and Power

The change in kinetic energy that occurs when work is done on an object is defined by the work-energy theorem:

$W = \Delta KE$

In this relationship, work ($W$) equals the change in kinetic energy ($\Delta KE$).

Power is defined as the amount of work done divided by the duration of time required to perform that work, representing the rate at which the energy of a system is altered by an external force:

$P = \frac{W}{\text{time}}$

While the base unit of measurement for power is the watt ($\text{W}$), the kilowatt ($\text{kW}$) is more frequently utilized because a single watt represents a very small quantity:

$1\,\text{kW} = 1000\,\text{W}$

Calculations and Application Examples

Example 1: Constant Horizontal Force

A soccer player exerts a constant $8.50\,\text{N}$ force over a distance of $11.8\,\text{m}$ on a $450\,\text{g}$ soccer ball across a soccer field. Determine the amount of work done by the player on the soccer ball.

Known quantities:

• $m = 450\,\text{g}$

• $F = 8.50\,\text{N}$

• $d = 11.8\,\text{m}$

Unknown quantity:

• $W = {?}$

Solution:

$W = F\,d$

$W = (8.5\,\text{N})(11.8\,\text{m})$

$W = 100.3\,\text{N}\cdot\text{m}$

$W = 100.3\,\text{J}$

Example 2: Force Exerted at an Angle

A husky pulls a sled across the ice over a distance of $50\,\text{m}$ using a strap oriented at a $25$-degree angle relative to the horizontal. Determine the amount of work the husky performs on the sled when exerting a force of $325\,\text{N}$ along the strap.

Known quantities:

• $F = 325\,\text{N}$

• $d = 50\,\text{m}$

• $\Theta = 25^\circ$

Unknown quantity:

• $W = {?}$

Solution:

$W = F\,d\cos(\Theta)$

$W = (325\,\text{N})(50\,\text{m})(\cos(25^\circ))$

$W = 14727.5$

$W = 1.47 \times 10^4$

Guidelines for Solving Work Problems

When working through problems involving work calculations, follow these procedural steps:

• Sketch the physical scenario, clearly identifying the vector representing the force performing the work alongside the displacement vectors.

• After applying the formula to compute the numerical value of work, evaluate whether system energy increased or decreased. This assessment determines whether the calculated work is positive or negative.

Conservation of Mechanical Energy

Fundamental Concepts of Mechanical Energy

Energy is neither created nor destroyed; it can only be changed from one form to another.

Mechanical energy is the sum of the kinetic energy and gravitational potential energy in a system determined by the position of its parts. More broadly, mechanical energy constitutes kinetic energy plus gravitational and elastic energy.

The total mechanical energy of a system neither increases nor decreases in any process, provided that only conservative forces are acting on the system.

Mathematical Formulation

When no other forms of energy are present, mechanical energy is represented by the formula:

$E = KE + PE$

In this relationship:

• $E$ represents total mechanical energy.

• $KE$ represents kinetic energy.

• $PE$ represents potential energy.

Under conservation conditions between two states (point 1 and point 2):

$\text{Total mechanical energy at point 1} = \text{Total mechanical energy at point 2}$

$\frac{1}{2} m v_1^2 + m g y_1 = \frac{1}{2} m v_2^2 + m g y_2$

Where:

• $m$ is the mass of the object.

• $g$ is the acceleration due to gravity, valued at $9.8\,m/s^2$.

• $v_1$ and $v_2$ are the initial and final velocities.

• $y_1$ and $y_2$ are the initial and final vertical positions.

Sample Problem: Velocity of a Falling Rock

A rock begins from an initial position of $y_1 = h = 3.0\,m$ at rest ($v_1 = 0$). The objective is to determine the speed of the rock when it falls to a position $y_2 = 1.0\,m$ above the ground.

1. State the conservation of mechanical energy equation:

$\frac{1}{2} m v_1^2 + m g y_1 = \frac{1}{2} m v_2^2 + m g y_2$

2. Cancel mass ($m$) from each term and substitute $v_1 = 0$:

$g y_1 = \frac{1}{2} v_2^2 + g y_2$

3. Rearrange and solve for $v_2^2$:

$v_2^2 = 2 g (y_1 - y_2)$

4. Substitute the known numerical values:

$v_2^2 = 2 (9.8\,m/s^2) (3.0\,m - 1.0\,m)$

$v_2^2 = 39.2\,m^2/s^2$

5. Calculate the final velocity $v_2$:

$v_2 = \sqrt{39.2\,m^2/s^2}$

$v_2 = 6.3\,m/s$

The speed of the rock $1.0\,m$ above the ground is $6.3\,m/s$ downward.

Work Energy Theorem

Fundamentals of the Work-Energy Theorem

The change in the kinetic energy of an object is equal to the net work done on the object; this principle is known as the Work-Energy Theorem.

$\text{Work} = \Delta KE$

Work is equal to a change in kinetic energy. This relationship provides the change in kinetic energy; initial or final energies cannot be determined unless additional information is provided.

Dynamics of Work and Energy Transfer

• If a box is pushed downhill and friction acts on the box, the kinetic energy change is equal to the work done by the force which is the push without friction.

• Decreasing speed is also applicable to the work-energy theorem.

• When the box sliding down the inclined plane is stopped by human effort, the inclined plane works on the box.

• Applicable to all types of energy, as work transfers energy from place to place as well as changing form.

Example 1: Work Done on a Baseball and Kinetic Energy

Determination of the kinetic energy of a $145\,\text{g}$ ball thrown with a speed of $25\,\text{m/s}$, and the net work done on the ball to accelerate it to this speed from rest:

Kinetic energy formula: $KE = \frac{1}{2}mv^2$

Initial and final states: $KE_1 = 0$ $KE_2 = 45\,\text{J}$

Calculation after the throw: $KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.145\,\text{kg})(25\,\text{m/s})^2 = 45\,\text{J}$

$\text{Net work done} = 45\,\text{J}$

Example 2: Work Done on a Decelerating Car

A car traveling at $60\,\text{km/h}$ brakes to a stop within a distance of $20\,\text{m}$. If the car travels twice as fast at $120\,\text{km/h}$, the stopping distance is evaluated assuming the maximum braking force does not depend on speed.

Governing conditions:

• The stopping force $F$ is constant.

• The work needed to stop the car, $Fd$, is proportional to the distance traveled.

• Force $F$ and displacement $d$ act in opposite directions.

• The final speed of the car is zero.

Derivation: $W_{\text{net}} = Fd\cos(180^\circ) = -Fd$

$-Fd = \Delta KE = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = 0 - \frac{1}{2}mv_1^2$

Stopping distance $d$ increases with the square of the speed: $d \propto v^2$

When the speed of the car is doubled: $\text{Stopping distance} = (2)^2 = 4\text{ times as great} = 80\,\text{m}$

Linear and Angular Momentum

Angular Momentum

Angular momentum is the rotational inertia multiplied by rotational velocity. It describes rotational movement (velocity) and rotational inertia around a particular axis, such as a planet in orbit around the sun. It can be expressed as the product of speed, radial distance, rotation, and mass. Angular momentum is represented by the symbol $L$ and is a vector quantity possessing both direction and magnitude.

$\text{Angular momentum} = L = I w$

Conservation of Angular Momentum

No change of angular momentum can occur if there is no external torque (a force that produces or changes rotation) acting on the rotating system. This principle is known as the conservation of angular momentum. The angular momentum with no torque or having internal torque before an event is equal to the angular momentum at any time:

$L_1 = L_2$

An example of angular momentum is the solar system, which includes the sun and other planets or celestial bodies. The angular momentum of the system is conserved because no external torque acts upon the rotation to change it.

Linear Momentum

Linear momentum is the product of the mass and linear velocity of an object. It is a vector quantity that points in the same direction as the object's velocity.

$p = m v$

$p = \text{mass} \times \text{velocity}$

Collisions

The net momentum of colliding objects in a system remains the same (no change) before, after, and during a collision. The forces acting during the collision are internal forces interacting within the system. Within any collision, the net momentum before the collision is equal to the net momentum after the collision.

Elastic Collisions

In an elastic collision, kinetic energy does not change within the system, and objects rebound without the generation of heat or deformation. Examples include collisions between hard objects made of steel, glass, or hard plastic. For instance, two trains traveling toward one another at the same speed bound off one another upon colliding with no loss in speed.

Inelastic Collisions

In an inelastic collision, kinetic energy decreases because it is transferred to other forms, such as thermal energy, sound energy, or material deformation. Objects involved are typically composed of soft, sticky materials (such as clay) and may stick together during the collision.

Conservation of Momentum Calculation

Problem: A large metal ball with a mass of $5\,\text{kg}$ rolls at $2\,\text{m/s}$ and hits a cricket ball with a $1\,\text{kg}$ mass resting on the grass. The velocity of the large ball is determined using the conservation of momentum.

$\text{Net momentum before roll} = \text{net momentum after roll}$

$(5\,\text{kg})(2\,\text{m/s}) + (1\,\text{kg})(0\,\text{m/s}) = (5\,\text{kg} + 2\,\text{kg})v$

$10\,\text{kg}\,\text{m/s} = (7\,\text{kg})$

$v = \frac{10}{6}\,\text{m/s}$

The resulting velocity is in the same direction as the larger ball.

Collisions

Fundamentals of Collisions and Internal Forces

Collisions are defined by the net momentum of colliding objects in a system remaining the same before, during, and after the collision. The forces acting during a collision are internal forces interacting within the system. Within any collision, the net momentum before the collision equals the net momentum after the collision.

The total mechanical energy of a system, which comprises potential energy and kinetic energy, is conserved in a collision if no external forces are present.

Elastic and Inelastic Collisions

Collisions are categorized based on whether kinetic energy is conserved and how materials behave during impact:

• Elastic Collisions:

  • The kinetic energy of the system does not change.

  • Colliding objects rebound without the generation of heat or any deformation.

  • Characteristic examples include collisions between hard objects made of steel, glass, or hard plastic.

• Inelastic Collisions:

  • The kinetic energy of the system decreases, with some kinetic energy being converted into thermal energy.

  • Collisions result in deformation and the generation of heat.

  • Colliding objects are typically made of soft, sticky materials such as clay and may stick together during the collision.

Principles of Conservation

Conservation of Momentum

Conservation of momentum is a fundamental physics principle stating that the total momentum of a closed system remains constant if no external forces act on it.

Momentum ($p$) is defined as the product of an object's mass ($m$) and its velocity ($v$):

$p = m v$

In a closed system devoid of external forces, the initial momentum before an event equals the final momentum after the event:

$\Sigma p_{\text{initial}} = \Sigma p_{\text{final}}$

Conservation of Mechanical Energy

Conservation of mechanical energy states that the total mechanical energy of a system remains constant in the absence of non-conservative forces, such as friction or air resistance.

Mechanical energy is the sum of an object's potential energy (derived from its position or state) and kinetic energy (derived from its motion). In an isolated system without non-conservative forces, the initial mechanical energy equals the final mechanical energy:

$E_{\text{initial}} = E_{\text{final}}$

Worked Example: One-Dimensional Cart Collision

A $2\,kg$ cart moves at $10\,m/s$ toward a $5\,kg$ cart initially at rest. The carts collide, causing the $2\,kg$ cart to stop and the $5\,kg$ cart to move. Calculate the speed of the $5\,kg$ cart after the collision, in $m/s$.

Solution Steps

Because the $5\,kg$ cart starts at rest and the $2\,kg$ cart ends at rest, the total initial momentum is carried entirely by the $2\,kg$ cart, and the total final momentum is carried entirely by the $5\,kg$ cart:

$\text{Initial momentum} = \text{Final momentum}$

$m v_i = m v_f$

$2(10) = 5(v_f)$

$v_f = 4\,m/s$

The speed of the $5\,kg$ cart after the collision is $4\,m/s$.

Collisions in One and Two Dimensions

One-Dimensional Collisions

In a one-dimensional collision, both the initial and final velocities of the involved masses lie along a single line. Because all motion variables occur within a single dimension, momentum needs to be conserved in one direction only (such as up/down or left/right).

Two-Dimensional Collisions

A collision in two dimensions obeys the same fundamental rules as a collision in one dimension:

• The total momentum in each direction is always the same before and after the collision.

• The total kinetic energy is conserved before and after the collision if the collision is elastic.

• Particles before and after the collision may have velocities directed across two dimensions. Therefore, momentum must be conserved independently in both directions using the respective velocity components along the $x$-axis and $y$-axis (up/down and left/right).

Predicting Collision Outcomes

One-Dimensional Predictions

• Objects travel strictly along a straight line before and after the impact.

• Applying the conservation of momentum allows for the determination of unknown velocities before or after the collision.

• In elastic collisions, the conservation of mechanical energy provides the necessary equations to determine both kinetic energies and velocities.

Two-Dimensional Predictions

• Objects move in multiple directions, requiring vector component analysis.

• Momentum conservation must be applied separately along each coordinate axis ($x$-axis and $y$-axis) to solve for directional final velocity components.

• Conservation of energy remains applicable if the two-dimensional collision is elastic.

Summary of Collision Characteristics

• Elastic Collision: Kinetic energy remains constant; net momentum before the collision equals net momentum after the collision.

• Inelastic Collision: Kinetic energy decreases and converts in part to thermal energy.

• Dimensionality Considerations: Momentum is conserved in both one-dimensional and two-dimensional collisions. One-dimensional analysis only requires evaluating up/down or left/right motion, whereas two-dimensional analysis requires evaluating both up/down and left/right components simultaneously.