Lesson Explainer: Mechanical Work Physics • 9th Grade

In this explainer, we will learn how to calculate the work done by a force that acts on an object over a distance.

When a net force acts on an object, the object accelerates. While accelerating, an object must move some distance.

An object that accelerates changes its kinetic energy. The change in the kinetic energy of the object due to the force acting on it is called the work done on the object by the force.

Let us consider an object with a mass, π‘š, of 2 kg and an initial velocity, 𝑒, of zero. The initial kinetic energy of the object is zero.

A constant force of 1 N acts on the object to increase its velocity to a final value, 𝑣, of 10 m/s. The kinetic energy of the object at its final velocity is given by KEKEkgmsKEJ=12Γ—π‘šΓ—π‘£=12Γ—2Γ—(10/)=100.

The displacement, 𝑠, of the object due to the force is given by 𝑣=𝑒+2π‘Žπ‘ , where the acceleration, π‘Ž, of the object is given by π‘Ž=12π‘Ž=0.5/.Nkgms

We see then that ο€»10=ο€»0+2Γ—0.5Γ—π‘ π‘š100=1Γ—π‘ π‘šο€Ό100οˆο€»1=π‘ π‘šπ‘ =100.msmsmsmsmsmmsms

We see that a force of 1 N acting over a distance of 100 m increases the energy of the object from zero to 100 J.

We can then define the work done by a force on an object as follows.

Definition: The Work Done on an Object by a Force

The work done, π‘Š, on an object by a force, 𝐹, is the product of the force and the distance moved, 𝑑, by the object in a direction parallel to the direction of the force. This can be written as π‘Š=𝐹𝑑.

If 𝐹 has the unit newtons and 𝑑 has the unit metres, then π‘Š has the unit joules.

It is important to recognize that the product of 𝐹 and 𝑑 is the change in the energy of the object, not the change in the velocity of the object.

We can see that the SI unit of work is the joule, which is also the SI unit of energy. This is to be expected as the work done on an object equals the change in the energy of the object. Just like energy, work is a scalar quantity.

For an object that moves in a single direction in a straight line, the distance moved by the object in the direction of the force is equivalent to the displacement of the object. It is important to understand that the displacement of the object has a direction and that the relation of the direction of the displacement to the direction of the force affects the work done by the force.

The following figure shows three objects that are all acted on by forces of equal magnitude. Each object is displaced equally while the forces act.

For the green object and for the blue object, the direction of the force and the direction of the displacement are parallel. For the orange object, the direction of the force and the direction of the displacement are perpendicular.

The distance moved, 𝑑, in the formula for work, π‘Š=𝐹𝑑, is a distance that must be parallel to the direction of the force. This is most easily interpreted for an object that moves in a straight line in one direction.

For the green object, as the force acts in the same direction as the displacement of the object, the force increases the kinetic energy of the object by 𝐹𝑑.

For the blue object, as the force acts in the opposite direction to the displacement of the object, the force decreases the kinetic energy of the object by 𝐹𝑑.

For the orange object, the force acts in a perpendicular direction to the displacement of the object. A force cannot accelerate an object in a direction that is perpendicular to the direction of the force. The force would in fact do zero work on the orange object as the object has zero displacement in the direction of the force. The kinetic energy of the object would not change.

For the orange object not to accelerate in the direction of 𝐹, something would have to be blocking the motion of the object in that direction, as shown in the following figure.

It is important to notice that the smooth surface blocking the motion of the object perpendicular to 𝑑 exerts a normal reaction force on the object that has the same magnitude and opposite direction to the applied force on the object. The net force on the object perpendicular to 𝑑 is therefore zero, as shown in the following figure.

The fact that the net force on the object is zero explains why its kinetic energy does not change.

A force may have a direction that is neither parallel nor perpendicular to the motion of an object, as shown in the following figure.

In this case, the force must be decomposed into a component 𝐹 parallel to 𝑑 and a component 𝐹 perpendicular to 𝑑, as shown in the following figure.

The work done on the object will equal the product of 𝑑 and the component of 𝐹 parallel to 𝑑, as shown in the following figure.

A force on an object and the displacement of the object can be described as being parallel if both the force and the direction of motion of the object act either in the same direction or in opposite directions throughout the motion of the object and the object does not reverse its direction of motion. This is shown in the following figure, in which a force acts either in the same direction as the motion of the object or in the opposite direction. The red arrows show the displacements of the object and the blue arrows show the force acting on the object at different object positions.

An object can change direction while it moves. The direction of the force that acts on the object can also change as the object moves, as shown in the following figure. The red arrow shows the path of the object and the blue arrows show the force acting on the object at different object positions.

Throughout the motion of the object, the direction of the force and the direction of the motion of the object are the same. The displacement of the object is shown by the black arrow, but the red curve shows the distance moved by the object in the direction of the force.

In the two preceding examples, if the force had the same constant value throughout the motion of the object and the lengths of the red lines were equal, then equal work would be done in each case. In both cases, the object would increase in speed as it moved.

An object can follow a curved path due to a force that acts perpendicularly to the motion of the object throughout its motion, as shown in the following figure. The red arrow shows the path of the object and the blue arrows show the force acting on the object at different object positions.

In this case, the force does no work on the object as the force never acts in the direction of the motion of the object. The object moves at a constant speed.

We may see examples in which a force does work on an object that moves a distance, in which the force does not have a stated direction and the direction in which the object moves is also not stated. In such examples, the directions of the force and the displacement can be assumed to be parallel throughout the motion of the object if no other information is given that could indicate otherwise.

Let us look at an example in which work is done by a force.

Example 1: Determining the Work Done by a Force

How much work is done by a 12 N force acting on an object and moving it a distance of 1.5 m?

Answer

Neither the force nor the distance have stated directions. To be able to answer this question, we must assume the relationship between the direction of the force and the direction in which the object moves.

The simplest assumption is that the motion of the object is parallel to the direction of the force throughout the motion of the object.

Neither the force nor the distance have stated directions, so it is not unreasonable to assume that they are parallel.

Making this assumption, we can use the formula π‘Š=𝐹𝑑 and substitute the values of 𝐹 and 𝑑 to obtain π‘Š=12Γ—1.5=18.NmJ

This is the work done on the object by the force, which equals the change in the kinetic energy of the object. This could be either an increase or a decrease in kinetic energy, as the force could act in either the same direction as or the opposite direction to the displacement of the object. In either case, the work done by the force is the same.

If the directions of the force and the displacement are not assumed to be parallel, the work done by the force on the object cannot be determined.

Let us now look at a similar example.

Example 2: Determining the Magnitude of a Force That Does Work

A force does 480 J of work pushing an object 32 m. What is the magnitude of the force?

Answer

Neither the force nor the distance have stated directions. The simplest assumption is that the motion of the object is parallel to the direction of the force throughout the motion of the object.

Making this assumption, we can use the formula π‘Š=𝐹𝑑.

We are asked to determine the magnitude of the force, so the formula must be rearranged to make 𝐹 the subject. This can be done by dividing the formula by 𝑑: π‘Šπ‘‘=𝐹𝑑𝑑=𝐹.

Substituting the values of π‘Š and 𝑑, we obtain 𝐹=48032=15.JmN

Let us now look at another similar example.

Example 3: Determining the Distance Moved by an Object That Work Is Done On

2β€Žβ€‰β€Ž240 J of work is done to a bookcase being pushed by a constant force of 1β€Žβ€‰β€Ž600 N. How far is the bookcase pushed by the force?

Answer

To answer the question, we must make the assumption that the motion of the object is parallel to the direction of the force acting on the object throughout the motion of the object.

Making this assumption, we can use the formula π‘Š=𝐹𝑑.

We are asked to determine the distance moved by the bookcase in the direction of the force, so the formula must be rearranged to make 𝑑 the subject. This can be done by dividing the formula by 𝐹: π‘ŠπΉ=𝐹𝑑𝐹=𝑑.

Substituting the values of π‘Š and 𝐹, we obtain 𝑑=22401600=1.4.JNm

Let us now look at an example in which a varying force does work on an object.

Example 4: Determining the Work Done by a Variable Force

The force applied to an object as it is moved over a distance is shown in the graph.

  1. What work is done to the object over the first 5 m that the object moves?
  2. What work is done to the object when moving from 5 m to 8 m away from the position of the object when the force was initially applied?

Answer

To answer the question, we must make the assumption that the motion of the object is parallel to the direction of the force acting on the object throughout the motion of the object.

Making this assumption, we can use the formula π‘Š=𝐹𝑑.

Part 1

For the part of the graph where the distance is between 0 metres and 5 metres, the force is not constant. The force is 0 newtons at a distance of 0 metres and is 100 newtons at a distance of 5 metres.

When a force changes uniformly, the maximum and minimum values of force can be used to determine the average force, which is given by 𝐹=100+02=50.averageNNN

We can see from the graph that the force increases by 20 newtons for each metre the object moves, meaning that the force changes uniformly.

The force changing uniformly tells us that using the average value of the force will give the same value of work done between 0 metres and 5 metres as if the value of the force at each position of the object was used to determine the work done at each position.

Using the formula π‘Š=𝐹𝑑, we obtain π‘Š=50Γ—5=250.NmJ

Part 2

The object moves a total of 8 metres. After it has moved 5 metres though, the force on the object is a constant 100 newtons. The average value of the force is now not required, and the constant value can be used.

Using the formula π‘Š=𝐹𝑑, we obtain π‘Š=100Γ—(8βˆ’5)=100Γ—3=300.NmNmJ

We have so far in this explainer only considered objects that are accelerated by forces. It is possible for an object to be acted on by forces and move at constant speed, provided that the forces are balanced.

Consider a box that is initially at rest on a smooth surface. A horizontal force acts constantly on the box and accelerates the box across the smooth surface. The smooth surface borders a rough surface. When the box moves onto the rough surface, it is moving at a nonzero velocity, 𝑣. The frictional force on the box due to the rough surface is equal in magnitude to the force pushing the box over the surface, so the forces on the box become balanced, and the velocity of the box is maintained at the value 𝑣 as the box moves over the rough surface.

This situation is represented in the following figure that shows the position of the box and the horizontal forces acting on it at equal time intervals. The blue arrows represent an applied force and the red arrows represent a frictional force.

While the box moves over the rough surface, it does not increase in velocity, and so it does not increase in kinetic energy. This means that the work done on the box is zero.

In this situation, the formula for work must have a value of zero for π‘Š, so we see that 0=𝐹𝑑.

Clearly, 𝑑 is nonzero, so 𝐹 must be zero. This is the case as the forces on the box are balanced and so the net force on the box is zero.

Suppose though that the box is being pushed by a person. The person is still pushing the box, and in pushing the box they are applying a force to the box while the box moves in the direction of the force.

The person is therefore doing work and, hence, is transferring energy. The person cannot though be transferring energy to the box, as no work is being done on the box.

So, if the person transfers energy, but not to the box, what happens to this energy?

This energy is said to be dissipated. Dissipation of energy is not the same thing as destroying energy; energy cannot be destroyed. Energy can be dissipated by sound and heat.

The physical laws required for describing dissipated energy in detail are not within the scope of this explainer, but the formula for work can be used to determine how much energy is dissipated when a force does work on an object.

Let us now look at an example in which dissipation of energy by a force is determined.

Example 5: Determining the Energy Dissipated by a Force That Does Work

An average force of 6β€Žβ€‰β€Ž250 N is applied by a car’s wheels while the car moves a distance of 880 m. The car’s engine does 10 megajoules of work to move the car. How much work, in megajoules, was done that did not involve moving the car, such as moving the air around the car, moving parts of the road surface, and heating parts of the car?

Answer

To answer the question, we must make the assumption that the motion of the car is parallel to the direction of the force on the car from the engine throughout the motion of the car.

Making this assumption, we can use the formula π‘Š=𝐹𝑑.

Substituting the values of 𝐹 and 𝑑, we obtain π‘Š=6250Γ—880=5500000=5.5Γ—10.NmJ

The conversion factor from joules (J) to megajoules (MJ) is given by 1Γ—=110Γ—=10.JMJMJ

The work done on the car by the engine in megajoules is therefore given by 5.5Γ—10Γ—10=5.5.MJ

The work done by the car’s engine is stated to be 10 MJ. Only 5.5 MJ of work was done to increase the kinetic energy of the car, however. The remaining energy transferred by the engine was dissipated. Dissipated energy includes such things as the energy of motion of the air, parts of the road surface, and the heating of parts of the car. The energy dissipated is given by 10βˆ’5.5=4.5.MJMJMJ

Let us now summarize what has been learned in these examples.

Key Points

  • The work done, π‘Š, on an object by a force, 𝐹, is the product of the force and the distance moved, 𝑑, by the object in a direction parallel to the direction of the force. This can be written as π‘Š=𝐹𝑑.
  • If 𝐹 has the unit newtons and 𝑑 has the unit metres, then π‘Š has the unit joules.
  • The work done on an object is equal to the change in the kinetic energy of the object.
  • For an object moving along a straight line in a single direction, where the force on the object acts parallel to the direction in which the object moves, the distance moved by the object in the direction of the force equals the displacement of the object, taking into account the signs of the force and the displacement.
  • Only the component of a force that acts in a direction parallel to the direction of the motion of an object does work on the object.
  • If a force acts in a direction perpendicular to the direction of the motion of the object, the force does no work on the object.
  • For an object that changes direction, if the force also changes direction to act in the same direction as the motion of the object throughout the motion, the force will do work on the object proportional to the distance moved by the object.
  • If an object is not accelerated by a force that acts on it, the work done by the force is not done on the object and is instead dissipated.

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