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

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Video Transcript

A child with a mass of 36 kilograms carries a sled with a mass of 14 kilograms to the top of an evenly sloping hill, walking 33 metres along the hillside and moving vertically upward by 8.8 metres. 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 metres per second when it arrives at the base of the slope. How much energy is dissipated during the sled’s downhill motion? What average force of friction does the hillside apply to the sled during the sled’s motion? Answer to the nearest newton.

Okay, we have here an example of energy conversion as well as energy conservation. Let’s clear some space on screen to draw a sketch of what’s going on. In this example, there is a child, a sled, and an evenly sloping hill. We’re told that the child walks to the top of the slope carrying the sled with them and then hops on top of the sled which is just enough to overcome the force of friction keeping the sled in place so that it all starts to slide down the hill. By the time they reach the bottom, we’re told that the child and the sled have a certain speed, 10 metres per second. Along with this, we’re also told what the mass of the child is. We can call it 𝑚 sub 𝑐 and that’s 36 kilograms. Furthermore, we’re told the mass of the sled. We’ve called it 𝑚 sub 𝑠 and that’s 14 kilograms.

Knowing all this, there are two questions that we want to answer. The first question is what is the energy dissipated during the descent. In other words, if we calculate the difference in the energy of our child and sled system at the top of the hill compared to the energy at the bottom, there will be some difference between them which is the energy dissipated by the sled and child while they sled down the hill. The second question is related to this. It is, “what is the average force of friction on the sled as it slides?” We’ll find that solving the first question will help us with the second. So let’s look at that one now, the energy dissipated during descent.

To calculate the energy dissipated, let’s establish two different points on our diagram. We’ll call them the initial and final points of the scenario. We’ll put the initial point up at the very top of our slope. We’ll say that at the point where the child and the sled are there, that’s the initial energy of our system. And then down at the bottom of the slope, we’ll put the final point. We’ll say that the energy of our child and sled system there is the final energy of the system. What we’re going to do here is rely on the principle of energy conservation. Recall that this principle says that in a closed system, the initial system energy is equal to the final system energy. Applying this principle to our scenario, our first task then is to figure out just what is the initial energy of our system and then what’s the final energy of the system.

To see what those are, let’s consider our initial and final points on our diagram. At the outset, the child we know is standing at the very top of the slope motionless with the sled. Since neither of them is moving, that means the initial energy of our system can’t include any kinetic energy or energy of motion. But at this initial position, we know that both the child and the sled are 8.8 metres above ground level. That means that they do both have gravitational potential energy. That sums up the initial energy of our system, the energy of the child and the energy of the sled.

We can recall that the gravitational potential energy of a mass of object 𝑚 is equal to that mass times the acceleration due to gravity times its height above some reference level. This means that in the case of our child and our sled, we can write that the quantity of the mass of the child plus the mass of the sled that is the total mass of our system multiplied by the acceleration due to gravity multiplied by the height 8.8 metres is equal to the final energy of the system. And we know further that the acceleration due to gravity near Earth’s surface can be approximated as 9.8 metres per second squared. Substituting in for that value as well as the mass of the child and the sled, we now have a left-hand side of our equation which fully expresses the initial energy of our system.

Now, we move on to the final state of our system, the point at which the child and the sled have just reached the bottom of the slope. At this location, both the child and the sled have returned to ground level so they no longer possess any gravitational potential energy. But we’re told that they are in motion, both having a speed of 10 metres per second. So this means 𝐸 sub 𝑓, the final energy of this system, at least consists of kinetic energy. And our question gives us a clue that it includes something else too. Our question refers to an amount of energy that’s dissipated during the descent. In other words, this is energy used up by friction between the sled and the slope. That energy goes into heating both the sled and the slope. And we consider it dissipated energy.

So we can write the final energy of our system as the sum of two terms. It’s the sum of the kinetic energy of the child and sled when they’re at the bottom of the slope plus the energy dissipated as the sled went down the slope. We can recall that an object’s kinetic energy is equal to one-half its mass times its speed squared. In our case, our object — so to speak — is the combination of the child and the sled. And we can see from our potential energy term that their combined mass is 50 kilograms. We fill in those terms for kinetic energy. And now, we have an equation which says that the potential energy of the child and sled when they’re at the top of the slope is equal to their kinetic energy when they’re at the bottom plus the energy dissipated. That’s the term we want to solve for.

To do that, let’s subtract the kinetic energy term from both sides of our equation. When we do that and we factor out the combined mass of our child and sled system, we get this expression for the energy dissipated. Entering the expression on our calculator, we find a result of 1812 joules. This is the amount of energy dissipated lost to friction over the sled’s descent. Like we said, this energy dissipated happened because of friction acting on the sled. In our second question, we want to solve for the average force of that friction.

To figure this out, we’ll need to recall the connection between work and energy. Recall that work is equal to force multiplied by distance. In our case, we can say that the work done by friction — we’ll call it 𝑊 sub 𝑓 — is equal to the force of friction on the sled multiplied by the distance the sled slides. As we said, it’s this force of friction which leads to the energy dissipation that we solved for in the first part. The frictional force between the sled and the slope is what led to this loss. And that amount of energy lost — the energy dissipated — what we’ve called 𝐸 sub 𝑑 is equal to the work done by friction.

We can think of it this way. As the sled slides down the slope, friction is constantly pushing in the opposite direction of the sled’s motion, trying to slow it down. All that force over all this distance adds up to work being done by friction which is equal to the energy dissipated as the sled descends. This means we can go back to our equation for the work done by friction and say that it’s also equal to the energy dissipated. This means we can solve for the average force of friction by rearranging this equation to solve for 𝐹 sub 𝑓. We know that 𝐸 sub 𝑑 is the value we solved for in part one. And 𝑑 in this case is 33 metres, the length of the incline. When we calculate this fraction, the result we find to the nearest newton is 55 newtons. That’s the average frictional force on the sled as it slides.

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