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In this lesson, we will learn how to apply the conservation of energy principle to solve problems on moving bodies.
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Letβs start by defining what conservation of energy is.
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Conservation of energy means that the total energy of an isolated system remains constant.
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In equation form, we can say that the total initial energy of the system, πΈ subscript π, is equal to the total final energy of the system, πΈ subscript π.
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In this video, when we talk about the total energy of the system, πΈ, we are talking about the potential energy plus the kinetic energy plus the work done against friction.
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We should recall that the gravitational potential energy of an object, ππΈ, is equal to ππβ, where π is the mass of the object measured in kilograms, π is the acceleration due to gravity measured in meters per second squared, and β is the height of the object above a reference point measured in meters.
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We also need to recall the kinetic energy of an object is equal to one-half ππ£ squared, with π once again representing the mass measured in kilograms and π£ being the speed of the object measured in meters per second.
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The work done against friction can be expanded out to be the force of friction times the displacement, where the force of friction is parallel to the displacement.
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The force of friction, πΉ subscript π, is measured in newtons and the displacement is measured in meters.
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A ball that is dropped will have all of its initial energy as potential energy when itβs dropped and all of its final energy as kinetic energy, just as itβs about to hit the ground.
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If air resistance is negligible, then we can say the potential energy of the ball when itβs dropped is equal to the kinetic energy of the ball just before it hits the ground.
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Letβs apply the conservation of energy principle to a few example problems, beginning with a body on a smooth inclined plane.
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A body started sliding down the line of greatest slope of a smooth, inclined plane.
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When it was at the top of the plane, its gravitational potential energy relative to the bottom of the plane was 1830.51 joules.
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When it reached the bottom of the plane, its speed was 8.6 meters per second.
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Find the mass of the body.
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When our object is at the top of the incline, it has a potential energy of 1830.51 joules.
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When it gets to the bottom of the inclined plane, itβs traveling at a speed of 8.6 meters per second.
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We can apply the principle of conservation of energy to determine our unknown mass.
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In equation form, the principle states that the total initial energy of a system is equal to the total final energy of the system.
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Looking back at our diagram, we can see that our initial energy is all potential energy.
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And since the object is moving when it reaches the bottom of the inclined plane, the total final energy is kinetic energy.
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To find the mass, we need to expand out the kinetic energy.
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Recall that kinetic energy is equal to one-half ππ£ squared, where π is the mass of the object and π£ is the speed of the object.
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Substituting in the expanded form for the kinetic energy, we now have the potential energy at the top of the incline is equal to one-half ππ£ squared of the object at the bottom of the incline.
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Plugging in the values given to us in the problem, we have a potential energy of 1830.51 and a speed of 8.6.
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To isolate the π, we multiply both sides by two.
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This cancels out the one-half on the right-hand side of the equation.
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Next, we divide both sides by 8.6 squared.
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This cancels out that term on the right side of the equation.
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This leaves us with only the term π on the right side of the equation.
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When we perform the calculations on the left side of the equation, we come up with 49.5.
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We are solving for the mass, which is measured in kilograms.
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The mass of the body that slid down the inclined plane is 49.5 kilograms.
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In the next problem, we will use the principle of conservation of energy to calculate the work done against friction on a rough inclined plane.
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A body was projected up a rough inclined plane from its bottom.
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Its initial kinetic energy was 242 joules.
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The body continued moving until it reached its maximum height and then slid back down to the bottom.
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When it reached the bottom, its kinetic energy was 186 joules.
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Find the work done against friction π during the ascent and the gain in gravitational potential energy π when the body was at its maximum height.
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Below, we have drawn a diagram to represent the problem.
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In the diagram, we have labeled the initial and final kinetic energy of the object at the bottom of the inclined plane.
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We have also labeled the unknown work done against friction for the ascent and the unknown gravitational potential energy of the body at its maximum height.
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We have two unknown variables that we will be solving for.
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Letβs start by solving for the work done against friction during the ascent.
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We can apply the conservation of energy principle, which states that the total initial energy of a system is equal to the total final energy of the system.
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Initially, the only energy the object has is kinetic energy.
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So we can replace the total initial energy of the system with the initial kinetic energy.
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The total final energy of the system is gonna be the kinetic energy of the object at the bottom of the incline plus the total work done against friction.
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The total work done against friction is going to be twice the work done against friction on the ascent because our object goes up and then back down the inclined plane.
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This is why we use two π to represent the total work down against friction.
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Substituting in our values for initial kinetic energy and final kinetic energy, we are now ready to solve for π.
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We start by subtracting 186 from both sides of the equation, canceling out that term on the right-hand side.
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The left side of the equation becomes 56, and the right side of the equation becomes two π.
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Next, we divide both sides of the equation by two, canceling out the term on the right-hand side of the equation.
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Dividing 56 by two, we get 28.
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And work is measured in joules.
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The work done against friction during the ascent is 28 joules.
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We can once again use the conservation of energy principle to determine the gravitational potential energy of the body at its maximum height.
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The total initial energy of the system is still the initial kinetic energy of the object.
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But the final total energy of the system at the maximum height is the gravitational potential energy π plus the work done against friction on the ascent π.
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Substituting in the values for the total initial kinetic energy and the work done against friction during the ascent, we are ready to solve for π.
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To isolate π, we must subtract 28 from both sides of the equation, thereby canceling out the 28 on the right-hand side, giving us a value for the gravitational potential energy of 214 joules.
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The gain in gravitational potential energy π when the body is at its maximum height is 214 joules.
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For our next example, we will apply the conservation of energy to calculate the change in energy of a body in free fall.
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A body of mass nine kilograms fell vertically from a point 3.4 meters above the ground.
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At a certain moment, the speed of the body was 3.9 meters per second.
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Determine the change in the bodyβs gravitational potential energy from this point until it reached a point 68 centimeters above the ground.
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Take π equals 9.8 meters per second squared.
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Our object was dropped from a height of 3.4 meters.
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We are looking for the change in gravitational potential energy between when our object is at a speed of 3.9 meters per second and when our object is at a height of 68 centimeters above the ground.
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Before we can determine the change in gravitational potential energy of our object, we must first find its height above the ground when its speed is 3.9 meters per second.
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We can use the conservation of energy principle, the total initial energy of the system equals the total final energy of the system, to determine the unknown height.
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The total initial energy is potential energy as our object is dropped from a height above the ground.
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Our total final energy is made up of potential energy, as our object is at some unknown height above the ground, and also kinetic energy, as our object is now moving.
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We should recall that the potential energy of an object is equal to the mass of the object times the acceleration due to gravity times the height of the object above the ground.
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To differentiate between our two heights, we can call our initial height β subscript π and our final height β.
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To substitute in for the kinetic energy of the object, we need to remember that the kinetic energy is equal to one-half times the mass times the speed of the object squared.
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Looking at the expanded equation, we can see thereβs an π in each one of our terms.
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Since our π is nonzero in this problem, we can cancel it out of each of the terms.
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Substituting in the values from our problem, we have 9.8 for π, 3.4 for initial height, and 3.9 for our speed.
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The left side of the equation becomes 33.32 and the right side of the equation becomes 9.8 times β plus 7.605.
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To isolate β, we can subtract 7.605 from both sides of the equation, cancelling out the term on the right-hand side.
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Subtracting the term from the left side of the equation, we get 25.715.
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The final step to solve for β is dividing both sides by 9.8, canceling out the term on the right-hand side.
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The height of the object above the ground when its speed is 3.9 meters per second is 2.62398 meters.
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Now that we know the height, we can find the change in potential energy from when the objectβs at 2.62398 meters and at 68 centimeters above the ground.
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The change in potential energy, ΞππΈ, is equal to the final potential energy minus the initial potential energy.
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Gravitational potential energy is equal to ππβ.
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So we can replace ππΈ final with ππβ final and ππΈ initial with ππβ initial.
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We are looking for the change in gravitational potential energy from when the objectβs at a height of 2.62398 meters and when the object is at a height of 68 centimeters above the ground.
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Therefore, 2.62398 meters will be our initial height and 68 centimeters will be our final height.
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But there is one problem.
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Theyβre measured in different units, meters versus centimeters.
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We will, therefore, need to convert 68 centimeters into meters, using the conversion factor that one meter is equal to 100 centimeters.
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When we convert 68 centimeters to meters by dividing by 100, we get 0.68 meters.
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Both of the terms on the right side of the equation have an π times π in it.
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Therefore, we can factor out the term π times π and put it out front.
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Substituting the values from our problem with the mass of nine, π of 9.8, final height of 0.68, and initial height of 2.62398, we calculate the change of potential energy to be negative 171.459 joules, where the negative tells us that this is a loss in energy.
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The problem asked us to find the change of potential energy and does not require us to tell them whether itβs a gain or a loss.
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So we can apply our absolute value to both sides of the equation.
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The change in gravitational potential energy between the two positions is 171.459 joules.
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In our next example problem, we will be using the conservation of energy principle where we need to account for friction.
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The figure shows a body of mass one-quarter kilograms before it started to slide along the surface.
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The two surfaces π΄π΅ and πΆπ· are smooth.
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However, the horizontal plane π΅πΆ is rough, and its coefficient of kinetic friction is seven-tenths.
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If the body started moving from rest, find the distance that the body covered on π΅πΆ until it came to rest.
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Consider the acceleration due to gravity to be π equals 9.8 meters per second squared.
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We can add the labels of mass is equal one-quarter kilogram and π, the coefficient of kinetic friction, is equal to seven-tenths to the figure.
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We can apply the principle of conservation of energy, where the total initial energy of the system is equal to the total final energy of the system, to our problem.
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Initially, at position π΄, the object is at rest, so the only form of energy that the object has is gravitational potential energy.
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The object comes to rest somewhere along the plane of π΅πΆ.
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Therefore, the final energy does not have any potential energy or kinetic energy, only the work done against friction.
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We should recall that the gravitational potential energy of an object is equal to the mass of the object times acceleration due to gravity times the height of the object above the ground.
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We can substitute in ππβ for the potential energy in our formula.
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We should also recall that the work done by friction is equal to the force of friction times the distance traveled, which allows us to substitute in force of friction times distance for the work.
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The problem does not give us the force of friction, but it does give us the coefficient of kinetic friction.
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So we must use the definition of force of friction, which is the coefficient of kinetic friction times the normal reaction force.
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The normal reaction force is the force that a surface puts on an object.
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When an object is traveling along a horizontal surface as, it does in this problem, without any other vertical forces besides the force of gravity and the normal reaction force acting on it, then we can say the normal reaction force is equal to the force of gravity, where the force of gravity is the mass of the object times the acceleration due to gravity.
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Looking at our expanded equation, we can see that thereβs an π and a π on both sides of the equation.
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Since these are both nonzero numbers, we can cancel them out.
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Now we can plug in the values from our problem, four for the height and seven-tenths for the coefficient of kinetic friction.
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To isolate π, we can multiply both sides of the equation by 10 over seven.
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This will cancel out the seven-tenths on the right side.
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The left side of the equation multiplies out to be 40 over seven.
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The distance the body covered on π΅πΆ until it came to rest is 40 over seven meters.
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Letβs review the key points of the lesson.
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Key Points
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The conservation of energy principle, total energy of an isolated system remains constant.
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In equation form, we can say the conservation of energy principle is πΈ subscript π, the total initial energy of the system, is equal to πΈ subscript π, the total final energy of the system.
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When friction is present, the work done against friction must be included in the calculation.