Lesson Explainer: Energy Conversion and Conservation Physics • 9th Grade

In this explainer, we will learn how to convert different types of mechanical energy to and from each other and to recognize when mechanical energy is dissipated.

People often talk about different types of energy. However, for the rest of this explainer, we will instead refer to different categories of energy. Our reason for doing this is to help make it clear that these distinctions are a matter of convention rather than fundamental natural properties. There is nothing fundamental about how or where energy is stored. Our classification of different energy categories effectively just acts as an accounting system for our own convenience.

We should also be clear that energy is not any kind of physical substance; it is a quantity like length, and we do not think of long things as being β€œmade of length,” but rather we would say they have a certain value of length.

When talking about energy conversion and energy conservation, the two terms go hand in hand. Whenever energy is converted from one category of energy to another, then the total amount of energy is conserved. We should point out that we can also talk about energy getting transferred. Energy transfer is a more general term that includes conversion of energy between categories but may also be used to mean energy getting moved, or transferred, between different objects or locations.

To see what these concepts of energy conversion and conservation mean, let’s consider the situation in which a ball is released from some height above the ground.

From experience, we know that the ball will not just stay floating in the air at the position it was released. Instead, it will fall to the ground, getting faster and faster the further it falls, as shown in the diagram below.

We can explain this observation in terms of energy conversion and energy conservation.

The ball has some energy, known as gravitational potential energy (or GPE for short), as a result of being raised up from the ground.

The reason it has this energy is that, when the ball is raised, work is done to move it in the opposite direction to its weight. The weight force acting on an object of mass π‘š is equal to π‘šπ‘”, where 𝑔 is the acceleration due to gravity. This weight force acts downward, and so to raise the ball an upward force must be applied, as illustrated in the diagram below.

The work done on an object by a force is given by the product of the magnitude of the force and the distance moved by the object in the direction of the force. In other words, for a force of magnitude 𝐹 that causes an object to move a distance 𝑑, the work done, π‘Š, is given by π‘Š=𝐹𝑑.

Work has units of energy, and doing work on an object transfers energy to that object. In the case of raising a ball against the force of gravity, that energy is GPE.

In order to raise the ball against the force of gravity, the upward force applied must at least match the downward weight force π‘šπ‘”. So, if the ball is raised to a height β„Ž by a force of magnitude π‘šπ‘”, then the work done on the ball is equal to π‘šπ‘”Γ—β„Ž.

This work done transfers an amount of GPE to the ball equal to the amount of work done, so that the GPE of the ball is given by

GPE=π‘šπ‘”β„Ž.(1)

So, we know that, when held at a height β„Ž, the ball possesses a GPE equal to π‘šπ‘”β„Ž. Since this GPE is proportional to the height β„Ž, we see that if the ball is dropped, then, as it falls down, it will lose GPE.

However, we also observe that the speed of the ball increases as it falls. In our first diagram, this was indicated by the arrows representing the ball’s motion getting longer the closer it got to the ground. We can recall that an object with mass π‘š that is moving at a speed 𝑣 has a kinetic energy (or KE for short) given by KE=12π‘šπ‘£.

So, as the GPE of the ball decreases, the KE of the ball increases. In other words, the GPE gets converted to KE as the ball falls.

If the ball is stationary at the instant it is released, then, since the speed of the ball is zero, it has no kinetic energy. All of the energy of the ball is gravitational potential energy.

At the instant the ball reaches the ground, the height above the ground is zero and so the ball has no gravitational potential energy. Assuming no energy is dissipated due to air resistance, then at this point all of the GPE has been converted to KE.

We will come back to energy dissipation in more detail later. For now, we point out that energy dissipation due to air resistance would involve energy getting transferred from the ball to the surrounding air.

In this case of no dissipation, we can say that initialGPEfinalKE=.

This is an example of energy conservation.

In fact, this principle applies more generally, and we can define energy conservation as follows.

Definition: Energy Conservation

Energy cannot be created or destroyed. The total energy of a self-contained system remains constant.

When we talk about a system, we mean any collection of objects. When discussing energy transfer, it makes sense to define the system of objects to include any objects that may be involved in the energy transfer processes.

In the case of our example with the ball, the system would be made up of the ball plus Earth. In this case, we would be treating Earth as a fixed point such that its GPE and KE could not change. Earth is included in the system as it provides the ground level against which the ball’s GPE is measured.

A self-contained system means a system of objects in which none of these objects interacts with any objects outside the system.

When talking about energy conservation, we could almost see it as two principles rolled into one. Firstly, if we consider the universe as a whole, then the total energy of the universe remains constant. Secondly, if we pick any system of objects within this universe and isolate them from everything else such that they cannot transfer energy to or from anything outside of the system, then the total energy of that system of objects remains constant. We point out that actually these are just the same principleβ€”in the first case the universe is one giant self-contained system.

Within a self-contained system, energy may be converted between categories and transferred between different objects, but the overall amount of energy does not change.

In contrast, for a non-self-contained system, energy can be transferred between objects in the system and objects outside of that system. In this case, though energy is conserved by each transfer process, we cannot say that the total amount of energy in the system remains constant, since energy may be transferred in or out.

These two cases of a self-contained and a non-self-contained system are illustrated visually in the diagrams below.

In practice, we do not generally end up considering non-self-contained systems very often, since we can usually define our system to include all objects that may be involved in energy transfer processes.

What the principle of energy conservation means in practice is that whenever energy is converted between different categories or transferred between different objects, no energy can be lost or gained overall in the process. This holds true no matter which different objects or categories of energy we are considering.

We can formalize this statement into a principle of energy conversion and conservation.

Definition: Energy Conversion (in the Absence of Dissipation)

Whenever energy is converted from one category, which we will call category 1, to a second category, which we will call category 2, then theamountofenergylostascategory1theamountofenergygainedascategory2=.

This assumes that the energy initially stored as category 1 energy is converted entirely to category 2 energy and that no energy is dissipated in the process.

Let’s see how this works for the case we already considered: a ball of mass π‘š dropped from a height β„Ž above the ground. Imagine we know the value of the height β„Ž that the ball is dropped from and we want to work out how fast the ball is moving at the instant it hits the ground; we will call this speed 𝑣.

We said already that the GPE of the ball gets converted to KE as it falls, and that the initial GPE of the ball at a height β„Ž is equal to the final KE of the ball the instant it reaches the ground.

Let’s look at this in terms of our principle of energy conversion and conservation.

Strictly speaking, the way in which we have defined GPE is that it is the energy transferred to raise an object by an amount β„Ž, no matter what height it started out at. But, whenever we are measuring the amount of GPE that an object has, we need a reference against which we can measure the object’s height. Typically, we define this β€œreference height” at ground level, where β„Ž=0m, and say that the GPE at β„Ž=0m is equal to 0 J.

So, in our example with the ball, the instant it reaches the ground, its height is 0 m, so we can say that its GPE at this instant is 0 J. Then the amount of GPE lost by the ball as it falls is equal to the initial GPE that it had at a height β„Ž. Using equation (1), we see that this is equal to π‘šπ‘”β„Ž.

The ball was released from a stationary position, so initially it had zero KE. This means that the amount of KE gained by the ball as it falls is equal to the KE it has at the instant it reaches the ground. Using equation (1), this is equal to 12π‘šπ‘£οŠ¨.

Now, our principle of energy conversion and conservation tells us that the amount of energy lost as type 1, which is GPE, is equal to the amount of energy gained as type 2, which is KE.

This confirms our earlier assertion that initialGPEfinalKE=, which we can now recognize as a special case of GPElostKEgained=.

We should note that this is a special case because in this particular instance the amount of GPE lost is equal to the initial GPE. In other words, all of the ball’s initial GPE is lost by the time it reaches the ground.

Substituting in our expressions for the initial GPE and final KE, we have that π‘šπ‘”β„Ž=12π‘šπ‘£.

We note that the mass π‘š of the ball appears on both sides of the equation, so we can cancel this out. Physically, what this means is that the final speed 𝑣 of the ball is independent of its mass; if we drop two balls of different masses, then (ignoring any effects of air resistance) both balls will reach exactly the same speed as they fall.

Canceling the mass π‘š, we have π‘”β„Ž=12𝑣.

If we multiply both sides by 2 and then take the square root of both sides, we get √2π‘”β„Ž=𝑣.

In other words, we have used the principle of energy conversion and conservation to find an equation for the final speed 𝑣 of the ball in terms of its initial height.

Let’s now have a look at a specific example.

Example 1: Energy Conservation and Conversion

A car is initially at rest before it starts to roll along a downward-sloping road with its engine turned off. While rolling, the car’s velocity increased by 1.4 m/s. What vertically downward distance does the car travel? Gravity is the only force that acts on the car.

Answer

We are told in the question that we have a car that is initially at rest. This means that it begins with a velocity of 0 m/s. We are also told that the car rolls along a downward-sloping road, and its velocity increases by 1.4 m/s.

Let’s draw a sketch to show the situation.

We have the vertically downward distance moved by the car as β„Ž on our diagram; is what we are trying to find. It is important to be clear that the question is not asking us for the distance traveled by the car along the road but rather the vertical distance the car moves downward.

We can recall that the gravitational potential energy of an object is related to its height β„Ž. That is, GPE=π‘šπ‘”β„Ž, where π‘š is the mass of the object and 𝑔 is the acceleration due to gravity.

Since the car loses height as it rolls along this downward slope, we know that it will lose GPE. In fact, specifically, since its height decreases by an amount β„Ž, then we know that GPElost=π‘šπ‘”β„Ž.

Now, we do not actually know the mass π‘š of the car, but, as we will see, it turns out that we will not need to know this mass.

Our principle of energy conversion and conservation tells us that the total energy in the system remains constant and that the GPE lost by the car must be equal to an amount of energy converted to another category or transferred elsewhere.

In this case, we are told that gravity is the only force acting on the car. This means we do not have any effects of friction or air resistance, so we do not need to worry about any dissipation of energy.

So where does this GPE go?

Well, we can also recall that the kinetic energy of an object is defined as KE=12π‘šπ‘£, where π‘š is the mass of that object and 𝑣 is the object’s speed (which is simply the magnitude of the velocity).

In the case of the car, it is initially stationary with a velocity of 0 m/s, so it begins with KEJ=0.

As the car rolls, it gains some KE as it picks up speed. The KE of the car when 𝑣=1.4/ms will be given by the expression KEmsfinal=12π‘šΓ—(1.4/).

Since the car begins with a KE of 0 J and ends up with a KE of KEfinal, we know that KEgainedKEms==12π‘šΓ—(1.4/).final

Now, thinking back to our principle of energy conversion and conservation, we see that the initial GPE is getting converted into this KE, and we must have that GPElostKEgained=.

If we substitute in our expressions for the GPE lost and the KE gained, we get the following equation: π‘šπ‘”β„Ž=12π‘šΓ—(1.4/).ms

We said earlier that it would not matter that we do not know the mass π‘š of the car, and now we see why this is. Since π‘š appears on both sides of the equation, we can cancel this term out so that we get an expression that is independent of the mass: π‘”β„Ž=12Γ—(1.4/).ms

Dividing both sides of the equation 𝑔, we get an expression for the vertical distance β„Ž that the car moves downward: β„Ž=(1.4/)2𝑔.ms

If we substitute in that 𝑔=9.8/ms, then we get β„Ž=(1.4/)2Γ—9.8/.msms

Evaluating this expression gives us β„Ž=0.1.m

Thus, our answer to the question is that the car travels a distance of 0.1 m vertically downward.

Just as energy can be converted from GPE to KE, it can also be converted the other way around.

Let’s look at an example where KE gets converted into GPE.

Example 2: Mechanical Energy Conversion

A skateboarder rolls upward along a curved ramp, as shown in the diagram, and reaches a point that is a vertical height of 7.5 m above the base of the ramp.

  1. What is the skateboarder’s speed at the point that is 1.1 m vertically above the base of the ramp? Give your answer to one decimal place.
  2. What would the initial speed at the base of the ramp required for the skateboarder to reach the top of the ramp be? Give your answer to one decimal place.

Answer

Part 1

In this first part of the question, we are asked to find the speed of the skateboarder when they are at a height of 1.1 m. We will label this height β„ŽοŠ§, so that we have β„Ž=1.1m, and we will label the speed at this height as π‘£οŠ§.

We can see from the diagram that the skateboarder reaches a maximum height of 7.5 m, which we will call β„ŽοŠ¨. Since this is the maximum height they reach, we know that at this point their speed must instantaneously be 0 m/s, since at this height they are changing direction and are about to come back down the ramp. We will label this speed as π‘£οŠ¨, so we have 𝑣=0/ms.

Let’s add these labels to our diagram.

So, we know the height of the skateboarder at two positions on the ramp, and at one of these two positions we also know the speed.

We can recall that an object of mass π‘š raised to a height β„Ž has a gravitational potential energy given by GPE=π‘šπ‘”β„Ž, where 𝑔 is the acceleration due to gravity.

We can also recall that an object of mass π‘š moving with a speed 𝑣 has a kinetic energy given by KE=12π‘šπ‘£.

We can see that the skateboarder will gain GPE as they go up the ramp, since their value of height β„Ž will increase. We can also see that they will lose KE as they go up the ramp, as their speed decreases from π‘£οŠ§ to 𝑣=0/ms.

Since we are not told anything about dissipated energy, we can assume that we do not need to worry about the possibility of dissipation.

In this case, the principle of conversion and conservation of energy means that KElostGPEgained=.

The KE lost is equal to the KE at β„ŽοŠ§, when their speed is π‘£οŠ§, minus the KE at β„ŽοŠ¨, when their speed is π‘£οŠ¨.

We can write this as KElost=12π‘šπ‘£βˆ’12π‘šπ‘£, or, by factoring out the 12π‘š, KElost=12π‘šο€Ήπ‘£βˆ’π‘£ο….

The GPE gained is equal to the GPE at β„ŽοŠ¨ minus the GPE at β„ŽοŠ§, which we can write as GPEgained=π‘šπ‘”β„Žβˆ’π‘šπ‘”β„Ž, or, by factoring out the π‘šπ‘”, GPEgained=π‘šπ‘”(β„Žβˆ’β„Ž).

Since we know that the GPE gained is equal to the KE lost, we can equate these two expressions for the GPE gained by the skateboarder and the KE lost by the skateboarder: 12π‘šο€Ήπ‘£βˆ’π‘£ο…=π‘šπ‘”(β„Žβˆ’β„Ž).

Since the mass π‘š appears on both sides, we can cancel it out. This means our result does not depend on the mass of the skateboarder.

Canceling the mass π‘š gives us 12ο€Ήπ‘£βˆ’π‘£ο…=𝑔(β„Žβˆ’β„Ž).

In this equation, we know the values of the heights β„ŽοŠ§ and β„ŽοŠ¨, and we also know the value of the speed π‘£οŠ¨. We are trying to find the value of π‘£οŠ§, so we need to rearrange the equation to make π‘£οŠ§ the subject.

To do this we first multiply both sides of the equation by 2: π‘£βˆ’π‘£=2𝑔(β„Žβˆ’β„Ž).

Then, we add π‘£οŠ¨οŠ¨ to both sides: 𝑣=2𝑔(β„Žβˆ’β„Ž)+𝑣.

Finally, we take the square root of both sides of the equation: 𝑣=2𝑔(β„Žβˆ’β„Ž)+𝑣.

So, we now have an expression for the speed π‘£οŠ§. We simply need to put in our values for the terms on the right-hand side.

We can recall that the acceleration due to gravity is 𝑔=9.8/ms. And we have that β„Ž=7.5m, β„Ž=1.1m, and 𝑣=0/ms.

Substituting in these values gives us 𝑣=2Γ—9.8/Γ—(7.5βˆ’1.1)+(0/).msmmms

Evaluating the expression under the square root gives us 𝑣=√125.44/.ms

Evaluating the square root gives us 𝑣=11.2/.ms

So, our answer to the first part of the question is that the skateboarder’s speed when they are at a height of 1.1 m above the base of the ramp is equal to 11.2 m/s.

Part 2

In the second part of the question, we are asked what speed the skateboarder would need to have at the base of the ramp in order to reach the top of the ramp.

As with the first part, we can solve this using our principle of energy conversion and conservation.

At the base of the ramp, the height of the skateboarder is 0 m, and since GPE=π‘šπ‘”β„Ž, this means that the GPE of the skateboarder is equal to 0 J. All the energy of the skateboarder at this point is kinetic energy.

If the skateboarder just reaches the top of the ramp, then the speed at the top of the ramp must be 0 m/s, since this is the point at which they instantaneously stop and change direction. This means that at this point, since KE=12π‘šπ‘£οŠ¨, the KE of the skateboarder is 0 J. All of their energy is GPE.

As in the first part of the question, the KE lost by the skateboarder gets converted to GPE as they go up the ramp. And the principle of energy conversion and conservation tells us that KElostGPEgained=.

Looking at the diagram in the question, we can see that the height reached by the skateboarder at the top of the ramp is given by β„Ž=7.5+2.7=10.2.mmm

Since we said that the GPE at the base of the ramp was 0 J, then the GPE gained by the skateboarder is equal to the GPE they have at the top of the ramp. So, we can write that the GPE gained by the skateboarder is given by GPEgainedm=π‘šπ‘”Γ—10.2.

Then, from our principle of energy conversion and conservation, we know that this is equal to the KE lost.

But we also know that the final KE at the top of the ramp is 0 J, so the KE lost is equal to the intial KE at the base of the ramp. If we say that the skateboarder had a speed 𝑣 at the base of the ramp, then we can write their initial KE (which is equal to the KE lost) as KElost=12π‘šπ‘£.

Substituting in our expressions for the KE lost and the GPE gained into our equation for the principle of energy conversion and conservation, we have 12π‘šπ‘£=π‘šπ‘”Γ—10.2.m

Since we are trying to find the initial speed 𝑣, we need to rearrange this equation to make 𝑣 the subject. This is exactly the same rearrangement that we did in the first part of the question. We first note that we can cancel the mass on each side of the equation: 12𝑣=𝑔×10.2.m

Then, multiplying both sides by 2 and then taking the square root of both sides, we get 𝑣=√2𝑔×10.2.m

Finally, we just need to substitute in that the acceleration due to gravity is 𝑔=9.8/ms and evaluate the expression: 𝑣=√2Γ—9.8/Γ—10.2𝑣=√199.92/𝑣=14.139…/.msmmsms

Noticing that the question asked for this answer to be given to one decimal place, we have our answer to the second part of the question; that is, in order to reach the top of the ramp, the speed of the skateboarder at the base of the ramp must be equal to 14.1 m/s.

We said already that this principle of energy conversion and conservation applies no matter what categories of energy we are talking about.

There are many different categories of energy. These include, but are not limited to, thermal energy (the internal energy of an object due to the energy of its particles), elastic potential energy (the energy stored when an object such as a spring is stretched or squashed), and magnetic energy (the energy stored when repelling magnetic poles have been pushed together or attracting poles have been pulled apart).

Energy conservation applies whenever energy is converted between any of these categories as well as any others not listed. It also applies when work is done on an object and hence energy is transferred to that object; in this case, the principle of energy conservation means that the amount of work done on the object is equal to the amount of energy gained by the object. Or, more simply, workdoneenergytransferred=.

We can recall that the work done, π‘Š, on an object by a force of magnitude 𝐹 is given by π‘Š=𝐹𝑑, where 𝑑 is the distance moved by the object in the direction of the force.

If we have a force acting in the direction of an object’s velocity, this force will do work in order to increase the kinetic energy of the object. In contrast, if we have a force such as friction acting in the opposite direction to the object’s motion, the work done by the force will reduce the object’s velocity and hence also reduce its KE. This second case is an example of dissipation of energy.

We already briefly mentioned that energy could get dissipated earlier in this explainer, when we were discussing a ball dropped from a height β„Ž. Let’s now look in a bit more detail at what this means.

Essentially, dissipation simply refers to any energy that is transferred to an object outside of the system. Typically, this energy is transferred as heat or sound, often to the surroundings such as the ground or the air. The reason we use the word dissipated when describing this energy is that energy transferred out of the system as heat or sound does not generally change into β€œuseful” mechanical energy directly.

In the case of our falling ball, we made a reference to air resistance on the ball. This is a means by which energy can be dissipated, or transferred from the ball to the air as heat.

Another common method of dissipation of energy is friction when pushing an object across a rough surface, as illustrated in the diagram below.

In this case, as a result of the friction, energy is transferred from the object being pushed to the surroundings as heat and possibly also as sound (this is why we may hear a screeching noise when dragging a chair across the floor).

We can modify our principle of energy conversion and conservation to apply it to cases where energy gets dissipated.

Definition: Energy Conversion (in the Presence of Dissipation)

Whenever energy is converted from one category, which we will call category 1, to a second category, which we will call category 2, then, if energy is dissipated in the process, theamountofenergylostascategory1theamountofenergygainedascategory2theamountofenergydissipated=+

This assumes that, besides the amount of energy that gets dissipated, the energy initially stored as category 1 energy is converted entirely to category 2 energy.

We should make it clear that energy that is dissipated is not energy that is lost. Energy conservation means that energy cannot be created and it cannot be lost. Dissipated energy is simply energy that is transferred out of the system we are considering at the time.

Let’s finish by looking at an example in which there is work being done on an object and there is energy dissipation involved.

Example 3: Mechanical Energy Conversion

A child with a mass of 36 kg carries a sled with a mass of 14 kg to the top of an evenly sloping hill, walking 33 m along the hillside and moving vertically upward by 8.8 m. The child puts the sled on the slope where it is just held in place by friction and carefully climbs on board. The added weight of the child is just enough to start the sled moving and it slides down the hill, moving at 10 m/s when it arrives at the base of the slope.

  1. How much energy is dissipated during the sled’s downhill motion?
  2. What average force of friction does the hillside apply to the sled during the sled’s motion? Answer to the nearest newton.

Answer

Part 1

In this question we have a child, with a sled, who climbs up a hill and then slides back down on the sled. Once they have climbed up the hill, they are initially stationary (with a speed of 0 m/s) at a vertical height of β„Ž=8.8m.

The child and sled move a distance of 𝑑=33m along the slope as they slide down, returning to a height of 0 m at the base of the slope. At this point, they have reached a speed of 𝑣=10/ms.

Since the child and the sled move together, we can consider them as one object for the purpose of our calculations. The total mass of this object is equal to the mass of the child, which is 36 kg, plus the mass of the sled, which is 14 kg. If we call this total mass π‘š, then we have π‘š=36+14=50.kgkgkg

We can draw a sketch to show this information.

By climbing the hill, the child and sled gain some gravitational potential energy. Since the GPE of an object of mass π‘š at a height β„Ž is defined as GPE=π‘šπ‘”β„Ž, where 𝑔 is the acceleration due to gravity, then for our case of π‘š=50kg and β„Ž=8.8m, we have GPEkgm=50×𝑔×8.8.

We can also recall that the acceleration due to gravity is given by 𝑔=9.8/ms, which means that the GPE of the child and sled at the top of the hill is GPEkgmsmGPEkgmsJ=50Γ—9.8/Γ—8.8=4312β‹…/=4312.

As the child then slides down the hill on the sled, the child and sled lose this GPE as the height decreases. Some of this GPE is converted into kinetic energy (KE) as they pick up speed and some is dissipated as a result of friction.

Our principle of energy conversion and conservation tells us that GPElostKEgainedenergydissipated=+.

We can work out how much KE is gained, since we know that KE is defined as KE=12π‘šπ‘£οŠ¨ and that the mass π‘š is 50 kg, the initial speed is 0 m/s, and the final speed is 10 m/s.

As the initial speed is zero, the initial KE is 0 J, and hence the KE gained is equal to the final KE: KEgainedkgmsKEgainedJ=12Γ—50Γ—(10/)=2500.

Then, we can substitute in our calculated values for the GPE lost (4β€Žβ€‰β€Ž312 J) and the KE gained (2β€Žβ€‰β€Ž500 J) into our equation for the principle of energy conversion and conservation: 4312=2500+.JJenergydissipated

We want to rearrange this to make the energy dissipated the subject. To do this, we subtract 2β€Žβ€‰β€Ž500 J from each side. Then, (by also swapping the left- and right-hand sides of the equation over), we have energydissipatedJJenergydissipatedJ=4312βˆ’2500=1812.

Thus, our answer to the first part of the question is that the amount of energy dissipated is equal to 1β€Žβ€‰β€Ž812 J.

Part 2

The second part of the question asks us to find the average force of friction applied to the sled by the hillside during the sled’s motion.

We said in the first part that the energy dissipation happened because of friction acting on the sled. More precisely, the force of friction does work on the sled, and the amount of energy dissipated is equal to the work done by this friction.

If we draw the friction force on our diagram, we see that, as the sled slides down the slope, the friction force is constantly pushing in the opposite direction to the sled’s motion, trying to slow it down.

We can recall that the work done, π‘Š, by a force of average magnitude 𝐹 acting over a distance 𝑑 is given by π‘Š=𝐹𝑑.

We are trying to find the average force, 𝐹, so let’s rearrange this equation by dividing both sides by 𝑑 (and by also swapping the left- and right-hand sides over) to make 𝐹 the subject: 𝐹=π‘Šπ‘‘.

In this case, the distance 𝑑 is the distance traveled along the surface of the slope, since the friction is acting the whole time the sled moves along this distance. So we have 𝑑=33m.

We also know that π‘Š==1812.energydissipatedJ

Substituting these values into our equation for 𝐹, we have 𝐹=181233.Jm

Evaluating this expression, we have 𝐹=54.Μ‡9Μ‡0.N

Noting that the question asked for our answer to the nearest newton, we have that the average friction force acting on the sled during its motion is equal to 55 N.

Key Points

  • The principle of energy conservation tells us that energy cannot be created or destroyed.
  • Energy can be converted from one category of energy to another. It may also be transferred between different objects or locations. Whenever energy is converted or transferred in this way, the total amount of energy always remains constant. This holds no matter what categories of energy we are considering.
  • Dissipated energy is any energy that gets transferred to an object outside of the system we are considering. This energy is typically transferred as heat or sound, often to the surroundings.
  • If energy of category 1 gets converted to energy of category 2, then the amount of energy lost as category 1 must be equal to the amount of energy gained as category 2 plus the amount of energy dissipated (if any).

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