Lesson Explainer: Power | Nagwa Lesson Explainer: Power | Nagwa

Lesson Explainer: Power Mathematics

In this explainer, we will learn how to find the power of a constant force using the relation 𝑃=𝐹×𝑣.

Let us start by recalling the definition of work.

Definition: Work Done by a Force

When a constant force, 𝐹, acts on a body parallel to its displacement and moves it a distance, 𝑑, then the work done, π‘Š, by this force is π‘Š=𝐹×𝑑. The work done by a force is measured in joules (J), which is equivalent to Nβ‹…m and the work is calculated over a period of time in which the body travels a given distance.

As an example, imagine that a person pushes a box with a mass of 30 kg across a horizontal floor. The person exerts a force of 50 N on the box as they push it, and the box moves a distance of 6 m. The work done by the person is then 50Γ—6=300β‹…NmNm. Since work is an energy, this can also be expressed using joules, which is the standard unit of energy with the symbol J. Given that 1β‹…=1NmJ, the value for the work done is equal to 300 J.

Now consider how long it takes for the person to push the box across the floor. Imagine it takes the person 12 s to move the box 6 m. The person expends 300 J of energy in 12 s, which is equivalent to expending 25 J each second. We can say that the person expends energy at a rate of 25 J per second. This is an expression of power. Power is the amount of energy expended per second as a force acts. Let us formally define power.

Definition: Power

For a constant force, 𝐹, that acts on an object as the object moves a distance, 𝑑, in the direction of the force over a time, 𝑑, the power, 𝑃, is given by 𝑃=𝐹×𝑑𝑑.

Notice that in the fraction on the right-hand side of the above equation, there is a factor of 𝑑𝑑, which is the distance the object moves over the time it moves for. This is the average speed of the object, 𝑣. We can substitute 𝑑𝑑 with 𝑣 to get 𝑃=𝐹×𝑣.

The average power, as a force acts on an object as it moves a certain distance, is equal to the force on the object multiplied by the average speed of the object.

The standard unit for power is the watt, with the symbol W, where 1 W =1 J/s =1 Nβ‹…m/s, but there are other units that can be used for power, such as horsepower, with the symbol hp. There are several different definitions of horsepower, but in all of the following questions, 1=735hpW.

Let’s have a look at some examples.

Example 1: Finding the Power of an Engine

Given that a car’s maximum speed is 270 km/h and its engine generates a force of 96 kg-wt, determine the power of its engine.


This question gives us values in nonstandard units, so let’s start by converting the values into standard units.

There are 1β€Žβ€‰β€Ž000 metres in a kilometer, so 270/=270000/kmhmh. There are 3β€Žβ€‰β€Ž600 seconds in 1 hour, so we divide this value by 3β€Žβ€‰β€Ž600 to get the top speed of the car in metres per second: 270000/=75/mhms.

1 kilogram-weight is equal to 9.8 newtons, so 96=96Γ—9.8=940.8kg-wtNN.

We can now use the formula 𝑃=𝐹𝑣 to find the power of the engine when the car is moving at its top speed: 𝑃=940.8Γ—75/=70560.NmsW

Because we have used the SI units for force and speed, the resulting value for the power that we get is in the SI unit for power, the watt (which has the symbol W).

The question requires us to give our answer in horsepower, however. Recall that 1=735hpW; we get our answer by dividing 70β€Žβ€‰β€Ž560 by 735: 𝑃=70560=70560735=96.Whphp

Example 2: Finding the Maximum Speed

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


This question gives us values in nonstandard units, so let’s start by converting them. 1=735hpW so 187=187Γ—735=137445hpWW. Also 1=9.8kg-wtN so 374=374Γ—9.8=3665.2kg-wtNN.

The tractor engine is providing a power of 137β€Žβ€‰β€Ž445 W. The maximum speed of the tractor is a constant, hence when the tractor is moving at this speed, we know the forces in the system will be balanced. This suggests that the force provided by the engine of the tractor will be equal in magnitude to the friction force that it is pulling against. We can use the formula 𝑃=𝐹𝑣 to find 𝑣. First, we must rearrange the formula by dividing both sides by 𝐹: 𝑃𝐹=𝑣.

We can then substitute in the values we have for 𝑃 and 𝐹 to give us 𝑣: 𝑣=1374453665.2=37.5/.WNms

Since we used standard units, the value we get for 𝑣 is in metres per second. The question, however, requires us to give the answer in kilometres per hour. 1/=3.6/mskmh, so 𝑣=37.5Γ—3.6/=135/.kmhkmh

Example 3: Finding the Maximum Speed and Power

A car of mass 5 tonnes is moving along a straight horizontal road. The resistance to its motion is directly proportional to its speed. When the car is traveling at 78 km/h, it is equal to 40 kg-wt per ton of the car’s mass. Given that the maximum force of the engine is 300 kg-wt, determine the car’s maximum speed 𝑣 and the power 𝑃 at which its engine operates at this speed.


In this question, we must first figure out how fast the car is moving before determining the power provided by the engine when the car moves at this speed.

The maximum force that the engine can provide is 300 kg-wt. This means that once the resistance on the car reaches this magnitude, the car cannot go any faster because the engine cannot produce a net force on the car, so the car cannot accelerate beyond this speed.

When the car is traveling at a speed of 78 km/h, it experiences a resistance force with a magnitude of 40 kg-wt per ton of its mass. Since its mass is 5 tonnes, this resistance force in total is equal to 200 kg-wt.

Since the resistance force is directly proportional to the speed, there is always a constant ratio between the speed of the car and the resistance force. This means that if the speed increases by a certain factor, the resistance force increases by the same factor. If the resistance force increases from 200 kg-wt to 300 kg-wt, it increases by a factor of 300200=1.5. This must correspond to an increase in speed by the same factor: 78/Γ—1.5=117/kmhkmh. This is the maximum speed of the car.

Since we already know the force produced by the engine at this speed, we can use 𝑃=𝐹𝑣 to find the power provided by the engine.

First, let’s convert 𝐹 into standard units: 𝐹=300kg-wt, and there are 9.8 N in each kg-wt, so 𝐹=300Γ—9.8=2940NN.

Second, let’s convert 𝑣 into metres per second. The maximum speed is 𝑣=117/kmh, and there are 1β€Žβ€‰β€Ž000 metres in one kilometre, so 117/=117000/kmhmh. There are 3β€Žβ€‰β€Ž600 seconds in an hour, so this speed is the same as moving 117000Γ·3600metres in each second, or 32.5 m/s.

Now, let’s use the formula to find 𝑃: 𝑃=𝐹𝑣=2940Γ—32.5/=95550.NmsW

Then, we convert this value for the power into horsepower: 𝑃=95550Γ—1735=130.WhpWhp

So, the maximum speed of the car is 117 km/h, and the power provided by the engine is 130 hp.

Sometimes, a resistive force such as friction is not the only force that acts against the applied force. For objects on a slope, the object’s own weight can partly act against an applied force.

Recall that, for an object on a slope, the weight of the object can be resolved into components that are parallel and perpendicular to the slope. The diagram below shows an object with weight π‘Š at rest on a slope at an angle πœƒ to the ground.

The component of the weight that acts parallel to the slope is equal to π‘Šπœƒsin, and the component of the weight that acts perpendicular to the slope is equal to π‘Šπœƒcos.

Example 4: Determining the Increase in the Power of Motion on an Inclined Plane with Resistive Forces

A vehicle of mass 3 tonnes was moving at 51 km/h along a horizontal section of road. When it reached the bottom of a hill inclined to the horizontal at an angle whose sine is 0.5, it continued moving at the same speed up the road. Given that the resistance of the two sections of road is constant, determine the increase in the vehicle’s power to the nearest horsepower. Take the acceleration due to gravity 𝑔=9.8/ms.


This question is easier if we first draw a diagram of the scenario. We are not told the angle of the slope, but we are told its sine, which means that πœƒ=0.5=30arcsin∘. Now, let’s draw the diagram.

The vehicle moves at the same speed when it is going up the hill as when it is on the at ground. The friction between the vehicle and the ground is the same too. That means that the only extra power provided by the engine as it goes up the hill is the power needed to produce a force to counteract the component of the vehicle’s weight that acts down the slope.

We can express this algebraically. Let’s say that when the vehicle is on the ground, the power provided by the engine is π‘ƒοŠ§. This provides a force 𝐹, and the vehicle moves at speed π‘£οŠ§. We can also say that π‘ƒοŠ¨, 𝐹, and π‘£οŠ¨ are then the same quantities for when the car is moving up the hill. We know that 𝑣=π‘£οŠ¨οŠ§, so we will just call both of them 𝑣.

When the vehicle is on the ground, 𝑃=𝐹𝑣.

In this case, the vehicle is moving at a constant speed, so 𝐹 must just be equal to the friction force on the vehicle, which we will call 𝐹R.

When the vehicle is on the hill, 𝑃=𝐹𝑣.

In this case, 𝐹 must be equal to 𝐹R plus the downhill component of the vehicle’s weight. Let’s draw a diagram of the vehicle showing the parallel and perpendicular components of its weight.

The downhill component of the weight is equal to π‘šπ‘”πœƒsin. So 𝐹=𝐹+π‘šπ‘”πœƒ,𝑃=(𝐹+π‘šπ‘”πœƒ)𝑣,𝑃=𝐹𝑣+π‘šπ‘”π‘£πœƒ.RRRsinsinsin

The question asks us to find the increase in the power, which is equal to π‘ƒβˆ’π‘ƒ=𝐹𝑣+π‘šπ‘”π‘£πœƒβˆ’πΉπ‘£.RRsin

Two terms cancel on the right-hand side to give π‘ƒβˆ’π‘ƒ=π‘šπ‘”π‘£πœƒ.sin

We know that sinπœƒ=0.5 and 𝑔=9.8/ms. The mass, π‘š, is equal to 3 tonnes, which is equal to 3β€Žβ€‰β€Ž000 kg. The speed, 𝑣, is equal to 51 km/h, which is equal to 513.6/ms. We can substitute these values into the above formula: π‘ƒβˆ’π‘ƒ=3000Γ—9.8/Γ—513.6/Γ—0.5=208250.kgmsmsW

Then, we can convert this into horsepower: π‘ƒβˆ’π‘ƒ=208250Γ—1735β‰ˆ283.WhpWhp The increase in the vehicles power is 283 hp, to the nearest whole number.

Key Points

  • Power is the rate at which a force does work.
  • The power, 𝑃, provided by a constant force 𝐹 that moves an object at a speed 𝑣 parallel to the force is given by 𝑃=𝐹𝑣.

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