Question Video: Applying the Work and Energy Principle to Solve Problems | Nagwa Question Video: Applying the Work and Energy Principle to Solve Problems | Nagwa

Question Video: Applying the Work and Energy Principle to Solve Problems Mathematics

Some workers are loading boxes in the back of a truck. Each box has a mass of 75 kg. And the height of the truck is 1 m. Given that the average total power that the group can work at is 0.5 hp, calculate the number of boxes they can load into the truck in a minute. Take 𝑔 = 9.8 m/sΒ².

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Video Transcript

Some workers are loading boxes in the back of a truck. Each box has a mass of 75 kilograms. And the height of the truck is one meter. Given that the average total power that the group can work at is 0.5 horsepower, calculate the number of boxes they can load into the truck in a minute. Take 𝑔 equals 9.8 meters per second squared.

We’ll call the mass of each box, 75 kilograms, π‘š. The height of the truck, one meter, we’ll call β„Ž. The average power of the group, 0.5 horsepower, we’ll call capital 𝑃. We want to solve for the number of boxes the group of workers can load into the truck in a minute. We’ll call that number capital 𝑁.

To start on our solution, let’s recall the mathematical relationship for power. Power 𝑃 is equal to the work done on an object divided by the time it takes to do that work. And speaking of work, we can further recall that this quantity is equal to force multiplied by distance. Combining these two equations for our situation, we can write that the power exerted by the group of workers 𝑃 is equal to the force they exert multiplied by the distance traveled over the time that this force is exerted.

The work they’re doing is lifting boxes of mass π‘š a height β„Ž into the back of a truck. And to do this, must exert a force on each box equal to the gravitational or weight force on the box of its mass times 𝑔. So in our equation, 𝐹 is equal to π‘š times 𝑔. And the distance 𝑑 is equal to β„Ž.

Currently, our equation expresses the power required to move a single box of mass π‘š. But we want to solve for the maximum number of boxes the workers can move in one minute. We can write this number 𝑁 as the total mass moved by the workers divided by the mass of a single box π‘š.

So if we rewrite our power equation to include the total mass moved as well as the total time allowed of 60 seconds, and we can replace π‘š sub 𝑑, the total mass of the boxes moved, by 𝑁, the total number multiplied by π‘š, the mass of a single box, we now have an equation which has the term we want to solve for as well as other terms which are given to us in the problem statement.

So we can rearrange to solve for 𝑁. We find that 𝑁 is equal to the power 𝑃 times 60 seconds divided by π‘š times 𝑔 times β„Ž, where π‘š is the mass of an individual box. When we plug these values in to solve for 𝑁, we include in the numerator a conversion factor to change our units of power from horsepower into watts.

When we enter these terms on our calculator, we find a value for 𝑁 just slightly above 30. But since boxes must come in whole numbers, the number 𝑁 rounds down to 30 boxes per minute. That’s the rate at which this group of workers is able to load boxes.

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