# Question Video: Calculating Power Based on Force and Speed Mathematics

A tractor has an engine of 187 hp and it is pulling against a force of 374 kg-wt. Find its maximum speed.

03:10

### Video Transcript

A tractor has an engine of 187 horsepower and it is pulling against a force of 374 kilograms-weight. Find its maximum speed.

All right, so let’s say that this is our tractor. And we’re told that it has an engine with a power of 187 horsepower. It exerts this power to pull against a force of 374 kilograms-weight. This is a way of telling us that the force opposing the tractor’s forward motion is 374 kilograms-weight. For the tractor to move forward as designed, it needs to match this force in the opposite direction. Its engine supplies the power to do that. The tractor will move at a constant speed when the forward force 𝐹 matches 374 kilograms-weight in magnitude.

And when all 187 horsepower of its engine are used, it will be moving at its maximum speed. These three quantities — power, force, and speed — are related by this equation. Since we want to solve for speed rather than power, we can divide both sides by the force 𝐹, cancelling that out on the right. This leaves us with the result that 𝑃 over 𝐹 equals 𝑣. In theory, we could solve for the maximum speed of the tractor by inputing our power and our force to this equation.

The trouble, though, is we would end up with units of horsepower per kilograms weight, an unfamiliar unit for speed. To express the maximum speed in more familiar units, we can convert horsepower and kilograms-weight into their respective SI base unit equivalents. One metric horsepower is equal to 735 watts. And one kilogram-weight equals 9.8 newtons. So if we multiply the numerator of our fraction by 735 watts per horsepower, then in terms of the units, horsepower will cancel out and we’ll be left with units of watts. And if we likewise multiply our denominator by 9.8 newtons per kilogram-weight, then kilogram-weight will cancel out, leaving us with units of newtons.

Note that in each case, as we multiplied by these conversion factors, we were effectively multiplying by one. That’s what makes these operations allowable algebraically. When all the dust settles then, we end up with units of watts per newton. And as we can infer from our equation, power divided by force is equal to speed, a watt per newton is equal to a meter per second.

We’re almost ready to calculate our maximum speed 𝑣. Before we do, there’s one last change to make. We could report 𝑣 in meters per second, but given the units we started with, horsepower and kilograms-weight, it’s more common to report speeds in kilometers per hour. Remembering that one meter per second is equal to 3.6 kilometers per hour, if we multiply our fraction by 3.6 kilometers per hour per meter per second, effectively multiplying by one, then we see the units of meters per second cancel from numerator and denominator.

Finally, then, we have an expression for 𝑣 in the units of interest. Computing the speed, we find a result of 135 kilometers per hour. This is the maximum speed the tractor can attain.