# Video: Determining the Relation of Work Done to Average Power Output

A person does 6.00 MJ of useful work in 2.88 × 10⁴ s. What is the person’s useful power output? How long will it take this person to lift 2000 kg of bricks to a 1.50-m-height platform?

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### Video Transcript

A person does 6.00 megajoules of useful work in 2.88 times 10 to the fourth seconds. What is the person’s useful power output? How long will it take this person to lift 2000 kilograms of bricks to a 1.50-metre-height platform?

As we work this problem, we’ll assume that the mass of bricks — the 2000 kilograms — is an exact number. And we’ll also assume that 𝑔 — the acceleration due to gravity — is exactly 9.8 metres per second squared. This is a two-part problem, where we’ll look first at the power output of this person. And then, we’ll find a time that it takes this person to perform a certain amount of work.

Let’s symbolize the power output we’re solving for with capital 𝑃 and then the time value we’ll solve for in part two with lowercase 𝑡. Now as we begin, let’s recall a relationship between power, work, and time. Power, symbolized using a capital 𝑃, is defined as an amount of work done divided by the time it takes to do that work. And the units of power is symbolized by a capital 𝑊, short for watts, where a watt is equal to a joule per second.

In our problem statement, we’re given an amount of work done by the person and we’re also given a time that it takes to do this work. So we have the ingredients necessary to compute the power involved. The person’s useful power output capital 𝑃 is equal to 6.00 megajoules divided by 2.88 times 10 to the fourth seconds.

We can rewrite the numerator of this equation in scientific notation 6.00 megajoules, where mega is the prefix signifying a million is equal to 6.00 times 10 to the sixth joules. Now, we have our equation for power in terms of units that will give us watts. When we divide this time value into this value of work which is in the same units as energy, we find that the power involved is 208 watts. That is this particular person’s. So that’s part one.

Now, we move on to part two to solve for the time it will take this person with this useful power output to lift 2000 kilograms of bricks to a 1.50-metre-high platform. Let’s write down that equation relating power, work, and time once more. We see that we have a time 𝑡 in this equation, which is just what we’re solving for. And the equation involves power 𝑃, which is what we solved for in part one.

Now, we can expand this equation because we call that work represented here by capital 𝑊 is equal to force times displacement. Knowing that, we can replace our 𝑊 with the force required to lift these bricks multiplied by the distance by which we need to move them. Now, as we look at this expanded form of the equation for power, we realize that the force we’re referring to — capital 𝐹 — is equal to the weight force that these bricks would exert. That that force is equal to the mass of the bricks given as 2000 kilograms times the acceleration due to gravity.

So let’s sub in 𝑚 times 𝑔 for capital 𝐹 in our equation for power. Now, as we look at this equation, the mass is known — that’s the mass of the bricks of 2000 kilograms. We also know 𝑔, the acceleration due to gravity, and 𝑑 — the displacement of the bricks — is given as 1.50 metres.

Since we already solved for the power in part one, all parts of this equation are known, except for 𝑡 — the variable we wanna solve for. So let’s rearrange this equation to solve for that time value. If we multiply both sides of the equation by 𝑡 — cancelling that term out on the right side — and then divide both sides of the equation by capital 𝑃 the power, cancelling that term out on the left side, we arrive at a final equation for 𝑡 which says that it equals the mass of the bricks times the acceleration due to gravity times the distance the bricks travel all divided by the person’s useful power output 𝑃.

Now, with this equation, we simply need to plug in the values of 𝑚, 𝑔, 𝑑, and 𝑃 that we we’ve either solved for, been given, or know as constant values. When we enter these numbers into our calculator — remembering that our mass 2000 kilograms and our acceleration due to gravity 9.8 metres per second squared are considered exact numbers — we end up with the time value of 141 seconds. That’s the amount of time it would take this person to lift 2000 kilograms of bricks to a 1.50-metre-high platform.