Guys, up until now, all of our problems have involved work and energy. In some problems, you'll need to calculate or find how quickly that energy gets transported or is done to another object. And that's exactly what power is. So, in this video, I'm just going to quickly introduce you to the equation for power. We're going to see what it is, and we're going to check out some problems. The equation for power states that the average power is equal to the amount of work done by an object divided by the change in time. It measures how quickly that work gets transferred to something else. This is often called average power because we calculate it between two points in time. Another way we can write this is by using the relationship between work and energy, as work is just a transfer of energy. Sometimes, you might actually see this equation written as ET. The unit we use for power is called watt, symbolized by 'W', which can be confusing. So, don't confuse this 'W' with the work done. Unfortunately, they both use the same letter, but a watt is really just a joule per second, representing energy divided by time.
We've actually seen this diagram before, which relates all of the important variables we've seen so far in the chapter like forces, works, and energies. Now, we are just adding another branch, another connection between work and power. As mentioned, power is about how quickly we are doing work, expressed as WtΔ. Let's go ahead and check out some examples.
Let's calculate how many joules of energy are used by a 100-watt light bulb. We have the power, P = 100, and we need to find out how much energy it uses in an hour. This is in hours, and we want it in seconds, so this will be 60 \text{ minutes} \times 60 \text{ seconds per minute}, which equals 3600 seconds. We have our average power equation, which is equal to the work done divided by time, and it's also related to the amount of energy consumed per time. We're trying to solve for 'E', which is E = P \cdot t, resulting in 100 \text{ watts} \times 3600 \text{ seconds} = 360,000 \text{ joules}. Most of your problems will be solved using this straightforward equation.
Moving on, let's consider moving objects, exerting forces, and changing velocities. Here we have a 1300 kg sports car on flat ground. It starts from rest, meaning that the initial velocity is 0. It is accelerating due to the engine's force, labeled Fengine. At a later time, the car's velocity reaches 40 m/s over a period Δt = 7 seconds. We need to figure out the average power delivered by the engine. This Paverage is calculated because the force of the engine moves the car through some distance, doing work in the process. To calculate this power, we consider the work done divided by the change in time or the energy divided by time. We know that work has been done, and with the change in time known as 7 seconds, we can determine the kinetic energy change: ΔKE = 12mvfinal2 - 12mvinitial2. The kinetic energy change lets us plug in this expression for our work to find the average power, leading to Paverage = 12mv2Δt, yielding 1.49 \times 10^5watts = 149 \text{ kilowatts}. You can convert this to horsepower if needed. This is how much power is delivered by the engine during this acceleration.
Hopefully, that made sense. Thank you for watching, and I'll see you in the next video.
