derive P = Fv and use it to solve problems

Energy Conservation in Physics

What is Energy Conservation?

Energy can neither be created nor destroyed; it can only change form. When a system is isolated, the total energy stays the same. Think of a closed box of marbles: you can roll them around, but you can’t magically add or remove marbles. This principle is the backbone of many physics problems, especially when we talk about power and work.

Power: The Rate of Energy Transfer

Power tells us how fast energy is being used or transferred. If you pour water into a bucket, the rate at which water fills the bucket is analogous to power. Mathematically, power is defined as the derivative of energy with respect to time:

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

In everyday life, we often measure power in watts (W), where 1 W = 1 J/s. A lightbulb that uses 60 W consumes 60 joules of energy every second.

Deriving the Formula $P = Fv$

Let’s derive the familiar power formula for a constant force acting on a moving object. We’ll walk through each step, using simple analogies to keep it clear.

1. Work done by a force: Work is the energy transferred when a force moves an object. If a force $F$ moves an object a distance $s$ in the direction of the force, the work is $$W = Fs.$$ Think of pushing a shopping cart: the harder you push (larger $F$) and the farther you push it (larger $s$), the more work you do.
2. Power as work over time: Power is the rate of doing work. If the work $W$ is done in a time interval $\Delta t$, the average power is $$P_{\text{avg}} = \frac{W}{\Delta t} = \frac{Fs}{\Delta t}.$$
3. Introduce velocity: Velocity $v$ is the distance per unit time, $v = \frac{s}{\Delta t}$. Substitute this into the power expression: $$P_{\text{avg}} = F \left(\frac{s}{\Delta t}\right) = Fv.$$ If the force and velocity are constant, this is the exact power.
4. Instantaneous power: For varying forces or velocities, we use calculus: $$P = \frac{dW}{dt} = \frac{d}{dt}(Fs) = F \frac{ds}{dt} + s \frac{dF}{dt}.$$ When $F$ is constant, the second term vanishes, leaving $P = Fv$.

So, whenever a constant force pushes an object at a steady speed, the power you’re using is simply the product of that force and the speed.

Example Problems

1. Car pulling a trailer: A car exerts a horizontal force of $500\,\text{N}$ to pull a trailer at a constant speed of $10\,\text{m/s}$. What is the power output of the car?

   $P = Fv = 500\,\text{N} \times 10\,\text{m/s} = 5000\,\text{W}$ (or 5 kW).

2. Elevator raising a load: An elevator lifts a 200 kg load at a constant speed of $1.5\,\text{m/s}$. Find the power required (ignore friction).

   Force needed = weight $= mg = 200\,\text{kg} \times 9.8\,\text{m/s}^2 = 1960\,\text{N}$. $P = Fv = 1960\,\text{N} \times 1.5\,\text{m/s} = 2940\,\text{W}$ (≈ 2.94 kW).

Practice Problems

1. A cyclist applies a force of $300\,\text{N}$ to maintain a speed of $8\,\text{m/s}$. What power is the cyclist delivering?
2. Water flows through a pipe at a rate of $0.02\,\text{m}^3/\text{s}$ and exerts a pressure of $5 \times 10^4\,\text{Pa}$. Calculate the hydraulic power.
3. A rocket engine exerts a thrust of $2 \times 10^6\,\text{N}$ while the rocket moves at $200\,\text{m/s}$. What is the power output?

Summary Table