14.3 Applications in Physics and Engineering

Physics and engineering use algebra to model everything from a car braking at a stoplight to current flowing through a circuit. Understanding these applications shows you why the algebraic techniques you've learned matter, and it gives you practice setting up and solving equations in context.

This section covers motion equations, forces and energy, projectile motion, work and power, simple harmonic motion, and electrical circuits and mechanical systems.

Modeling Physical Phenomena with Algebra

Motion Modeling

Motion equations connect four quantities: position ($x$), velocity ($v$), acceleration ($a$), and time ($t$). When acceleration is constant, three core equations let you solve for any unknown if you know enough of the others:

• $v = v_0 + at$
• $x = x_0 + v_0 t + \frac{1}{2}at^2$
• $v^2 = v_0^2 + 2a(x - x_0)$

Each equation leaves out one variable (position, time, or final velocity), so pick the equation that matches the information you have.

Example: A car starts from rest ($v_0 = 0$) and accelerates at $3 \text{ m/s}^2$ for $5$ seconds. To find how far it travels, use the second equation:

$x = 0 + 0(5) + \frac{1}{2}(3)(5^2) = 37.5 \text{ m}$

To find its final velocity, use the first equation:

$v = 0 + 3(5) = 15 \text{ m/s}$

Notice how choosing the right equation saves you from solving a system. That's the whole strategy: identify what you know, identify what you need, and pick the equation that connects them.

Force and Energy Modeling

Newton's Second Law ties force, mass, and acceleration together:

$F = ma$

This means if you know any two of the three quantities, you can solve for the third. For instance, pushing a $12 \text{ kg}$ box with a net force of $36 \text{ N}$ gives an acceleration of $a = \frac{36}{12} = 3 \text{ m/s}^2$.

Energy comes in two main forms for these problems:

• Kinetic energy: $KE = \frac{1}{2}mv^2$ (energy of motion)
• Potential energy: $PE = mgh$ for gravitational, $PE = \frac{1}{2}kx^2$ for elastic (springs)

Total mechanical energy is the sum: $TE = KE + PE$.

The conservation of energy principle says that if no energy is lost to friction or other non-conservative forces, total energy stays constant. Algebraically, you set the total energy at one point equal to the total energy at another:

$KE_1 + PE_1 = KE_2 + PE_2$

Example (roller coaster): At the top of a $20 \text{ m}$ hill, a $500 \text{ kg}$ cart moves at $4 \text{ m/s}$. At the bottom ($h = 0$), all that potential energy has converted to kinetic energy. Setting $\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + 0$, you can solve for $v_2$. The mass cancels, which is a useful pattern to watch for.

Solving Problems with Algebraic Techniques

Projectile Motion

Projectile motion splits into two independent parts: horizontal (constant velocity) and vertical (constant acceleration due to gravity). This separation is what makes the algebra manageable.

• Horizontal: $x = v_{0x} t$
• Vertical: $y = y_0 + v_{0y} t - \frac{1}{2}gt^2$
• Vertical velocity: $v_y = v_{0y} - gt$

Here $g \approx 9.8 \text{ m/s}^2$. The horizontal and vertical equations share the variable $t$, which is how you connect them.

Steps for solving a typical projectile problem:

1. Break the initial velocity into components: $v_{0x} = v_0 \cos\theta$ and $v_{0y} = v_0 \sin\theta$
2. Use the vertical equation to find the time of flight (often by setting $y = 0$ and solving the resulting quadratic)
3. Plug that time into the horizontal equation to find the range

Common pitfall: When solving $0 = v_{0y}t - \frac{1}{2}gt^2$, you'll get two solutions. One is $t = 0$ (the launch moment). The other is the landing time you actually want. Don't forget to factor out $t$ before solving.

Work and Power

Work measures energy transferred by a force over a distance:

$W = Fd\cos\theta$

The angle $\theta$ is between the force direction and the displacement direction. When force and displacement point the same way, $\theta = 0$ and $\cos\theta = 1$, so $W = Fd$. When the force is perpendicular to displacement (like gravity on an object moving horizontally), $\cos 90° = 0$ and no work is done.

Power is how fast work gets done:

$P = \frac{W}{t}$

Example: A $200 \text{ N}$ force pushes a crate $8 \text{ m}$ along the floor in $4 \text{ s}$, with force parallel to displacement. Work is $W = 200 \times 8 = 1600 \text{ J}$. Power is $P = \frac{1600}{4} = 400 \text{ W}$.

These formulas are straightforward, but the algebra gets interesting when you combine them with energy equations or need to solve for an unknown force or distance.

Simple Harmonic Motion with Trigonometry

Modeling Simple Harmonic Motion

Simple harmonic motion (SHM) occurs when a restoring force pulls an object back toward equilibrium, and that force is proportional to how far the object has been displaced:

$F = -kx$

The negative sign means the force always opposes the displacement. A stretched spring pulls back; a compressed spring pushes out.

Because this restoring force produces oscillation, the position, velocity, and acceleration are modeled with trig functions:

• $x(t) = A\cos(\omega t + \phi)$
• $v(t) = -A\omega\sin(\omega t + \phi)$
• $a(t) = -A\omega^2\cos(\omega t + \phi)$

Here $A$ is the amplitude (maximum displacement), $\omega$ is the angular frequency (in rad/s), and $\phi$ is the phase angle (which accounts for where in the cycle the motion starts).

Notice that the velocity equation is the derivative of position, and acceleration is the derivative of velocity. If you've seen derivatives in this course, that connection reinforces why these trig functions appear: the derivative of cosine is negative sine, and the derivative of sine is cosine.

Period and Frequency of Simple Harmonic Motion

The period $T$ is the time for one full oscillation, and the frequency $f$ is how many oscillations happen per second. They're reciprocals of each other:

$T = \frac{2\pi}{\omega}, \quad f = \frac{1}{T} = \frac{\omega}{2\pi}$

For a mass-spring system, the angular frequency depends on the spring constant and mass:

$\omega = \sqrt{\frac{k}{m}}$

Example: A $0.5 \text{ kg}$ mass on a spring with $k = 200 \text{ N/m}$ has $\omega = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20 \text{ rad/s}$. The period is $T = \frac{2\pi}{20} \approx 0.314 \text{ s}$, and the frequency is $f \approx 3.18 \text{ Hz}$.

A key takeaway: increasing the mass makes the oscillation slower (larger $T$), while a stiffer spring (larger $k$) makes it faster. That relationship is built right into the formula.

Applying Algebra to Circuits and Systems

Electrical Circuits

Circuit analysis is really just solving systems of equations, which makes it a natural fit for algebra.

Ohm's Law relates voltage, current, and resistance:

$V = IR$

Kirchhoff's Laws give you the equations to set up:

• Current Law (KCL): The total current entering any junction equals the total current leaving it.
• Voltage Law (KVL): The sum of all voltage gains and drops around any closed loop is zero.

For combining resistors:

• Series: Resistances add directly: $R_{total} = R_1 + R_2 + \cdots + R_n$
• Parallel: Reciprocals add: $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}$

Example (series): Three resistors of $4 \text{ Ω}$, $6 \text{ Ω}$, and $10 \text{ Ω}$ in series with a $40 \text{ V}$ battery. Total resistance is $20 \text{ Ω}$, so current is $I = \frac{40}{20} = 2 \text{ A}$.

Example (parallel): Two resistors of $6 \text{ Ω}$ and $3 \text{ Ω}$ in parallel give $\frac{1}{R_{total}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$, so $R_{total} = 2 \text{ Ω}$. For more complex circuits with multiple loops, you'll write KVL equations for each loop and solve the resulting system.

Mechanical Systems

Simple machines like levers, pulleys, and inclined planes trade off force and distance. Two key quantities describe their performance:

• Mechanical advantage (MA): $MA = \frac{F_{out}}{F_{in}}$. An MA of 3 means the machine multiplies your input force by 3, but you'll need to apply that force over 3 times the distance.
• Efficiency: $\eta = \frac{W_{out}}{W_{in}} \times 100\%$. Real machines always have efficiency below 100% because of friction.

Example: A lever lets you lift a $600 \text{ N}$ load by applying $200 \text{ N}$ of force. The mechanical advantage is $\frac{600}{200} = 3$. If you push down $0.9 \text{ m}$ and the load rises $0.25 \text{ m}$, the input work is $200 \times 0.9 = 180 \text{ J}$ and the output work is $600 \times 0.25 = 150 \text{ J}$. Efficiency is $\frac{150}{180} \times 100\% \approx 83.3\%$. The missing $30 \text{ J}$ went to friction.

These problems are algebraically straightforward, but they test whether you can identify which formula applies and correctly rearrange it to isolate the unknown.