Lesson Explainer: Conservation of Energy | Nagwa Lesson Explainer: Conservation of Energy | Nagwa

Lesson Explainer: Conservation of Energy Mathematics

In this explainer, we will learn how to apply the conservation of energy principle to solve problems on moving bodies.

The principle of the conservation of energy is relevant to such problems when we consider kinetic energy, gravitational potential energy, and energy dissipated due to friction.

Let us define kinetic energy.

Definition: Kinetic Energy

The kinetic energy of a body is dependent on the mass and velocity of the body, according to the formula 𝐾=12𝑚𝑣, where 𝑚 is the mass of the body and 𝑣 is the velocity of the body.

Let us define gravitational potential energy.

Definition: Gravitational Potential Energy

The gravitational potential energy of a body is dependent on the mass and altitude of the body, according to the formula 𝑃=𝑚𝑔, where 𝑚 is the mass of the body, is the altitude of the body, and 𝑔 is the acceleration due to gravity, which is typically taken to be 9.8 m/s2 near the surface of Earth.

A closed system is a set of bodies that transfer energy only between each other and not to any bodies not included in the system. In some questions, Earth may be considered one of the bodies included in the system. The sum of the potential and kinetic energy in a closed system is conserved and so is constant. Kinetic energy can be transferred to potential energy, and vice versa.

Energy can also transfer between systems. Such systems are termed open systems. One method of inter system energy transfer is where a body in one system does work on a body in a different system.

Let us define the work done on a body by a force.

Definition: Work Done on a Body by a Force

The work done on a body by a force is dependent on the force that acts on the body and the distance that the body moves in the direction of that force, according to the formula 𝑊=𝐹𝑑𝜃,cos where 𝐹 is the force, 𝑑 is the displacement of the body while the force acts on it, and 𝜃 is the angle between the directions of 𝐹 and 𝑑.

Forces can be defined as either conservative or non conservative.

A force is conservative if the work done on a body by the force depends only on the initial and final positions of the body and hence is independent of the path taken by the body during its motion. Gravitational forces are an example of conservative forces.

For a conservative force acting on a body, the work done by the force is related to the change in potential energy of the body. The relationship between the work done by the force and the change in the potential energy of the body is given by 𝑊=Δ𝑃, where 𝑊 is the work done by the force and Δ𝑃 is the change in the potential energy of the body.

The work–energy principle states that the work done on a body by a conservative or non conservative force equals the change in the kinetic energy of the object. This can be expressed as 𝑊=Δ𝐾, where 𝑊 is the work done by the force and Δ𝐾 is the change in the kinetic energy of the body.

We see then that for a conservative force acting on a body, Δ𝑃=Δ𝐾.

This can be expressed as Δ𝐾+Δ𝑃=0.

This can also be expressed as 𝐸=𝐾+𝑃, where 𝐸 is the energy of the system, which is a conserved quantity. The sum of the energies of open systems that interact is also a conserved quantity.

A force is non conservative if the work done on a body by the force depends on the path taken by the body during its motion. Frictional forces are an example of non conservative forces.

The work done by a non conservative force that acts on a body is not necessarily done on the body. Consider, for example, a horizontal force acting on a body on a rough horizontal plane. The action of such a force can result in the uniform motion of the body along the plane. During an interval in which the force acts on the body and the body moves uniformly, the force does work but the kinetic energy of the body does not change. The work done by the force that does not increase the energy of the body is said to be dissipated. A system in which energy is dissipated is a dissipative system. For a dissipative system, the energy of the system can be expressed as 𝐸=𝐾+𝑃+𝑊, where 𝑊 is the energy dissipated.

We can see that the principle of conservation of energy has many similarities to the work–energy principle. For non dissipative systems, these principles are equivalent.

Let us consider an example of energy conservation in a closed system.

Example 1: Calculating the Energy of a Body in Freefall

A body of mass 20 kg fell from a height of 42.3 m above the surface of the ground. Find the sum of its kinetic energy and its potential energy relative to the ground 2 seconds after it started falling. Take 𝑔=9.8/ms.

Answer

The body is described as falling from a height. This implies that the body is at rest before it starts to fall, so it has no kinetic energy initially.

The energy of the system when the body is at rest is given by its initial gravitational potential energy, 𝑃=20×9.8×42.3=8290.8,initialjoules as initially 𝐾 is zero.

When the body has fallen for a time 𝑡, it has decreased in altitude above the ground, and so the gravitational potential energy decreases by Δ𝑃: 𝑃=𝑃Δ𝑃.initial

The energy of the system is conserved, and so gravitational potential energy of the system is transferred to kinetic energy of the system. The transfer of energy to kinetic energy means that, at the instant 𝑡, 𝐾=0+Δ𝐾, where Δ𝐾=Δ𝑃.

The sum of 𝑃 and 𝐾 is therefore 𝑃+𝐾=𝑃Δ𝑃+Δ𝑃=𝑃=8290.8.initialinitialjoules

It is important to note that the time that the body fell for is not relevant to determining the sum of 𝑃 and 𝐾. Energy in a closed system is conserved, so 𝐸 for a closed system has the same value at all times.

Let us consider another example of transfer from gravitational potential energy to kinetic energy.

Example 2: Applying the Conservation of Energy Principle to Calculate Speed

A body started to slide down a smooth inclined plane of height 504 cm from its top. Find its speed when it reached the bottom. Take 𝑔=9.8/ms.

Answer

The body is described as starting to slide from a height. This implies that the body is at rest before it starts to slide, so it has no initial kinetic energy. The following figure represents the body at its initial and final positions.

The energy of the system when the body is at rest is given by its initial gravitational potential energy, 𝑃=𝑚×9.8×5.04,initial as initially 𝐾 is zero. The height in centimetres is converted to a height in metres to make it consistent with the unit m/s2 of 𝑔=9.8/ms.

The slope is smooth, so no dissipation of energy occurs. When the body is at the bottom of the slope, 𝑃 is zero. All the gravitational potential energy is transferred to kinetic energy. When this has happened, 𝐾=𝑃,initial so 12𝑚𝑣=𝑚×9.8×5.04.

There is a common factor of 𝑚 on both sides of the equation that can be eliminated to give 12𝑣=9.8×5.04,𝑣=2×495×12625,𝑣=2×495×12625=12348125𝑣=9898125.

Taking the square root of both sides gives 𝑣=42575𝑣=423525/.ms

It is important to note that the angle of the incline that the body slides along is not relevant to determining the energy of the system. The path that the body takes from its initial height to the bottom of the incline does not affect the decrease in 𝑃. The increase in 𝐾 due to the decrease in 𝑃 is the same for any path taken by the body.

Intuitively, it might be expected that a vertically falling body would acquire a greater speed than a body sliding down an incline from the same height as the falling body, due to the greater acceleration of the vertically falling body. This expectation is misleading however, as a body that slides along the incline accelerates for a greater time than the vertically falling body, and the increased acceleration time exactly compensates for the decreased acceleration.

Now let us look at an example where the speed of the body at a particular instant is considered.

Example 3: Calculating the Change in Potential Energy for a Body on a Smooth Inclined Plane

A particle of mass 281 g was projected at 37 cm/s up the line of greatest slope of a smooth plane inclined to the horizontal at an angle whose sine is 1011. Determine the change in the particle’s gravitational potential energy from the moment it was projected until its speed became 29 cm/s.

Answer

For the particle, initially, 𝑃=0. The sloping plane that the particle moves along is smooth, so no energy is dissipated, and the increase in 𝑃, Δ𝑃, is equal to the decrease in 𝐾, Δ𝐾.

The value of Δ𝐾 can be determined by subtracting the value of 𝐾 when the particle moves at 37 cm/s from the value of 𝐾 when the particle moves at 29 cm/s. This can be expressed as Δ𝐾=12(281)2937=74184.

This is not the value of Δ𝐾 in joules, however, as the mass was not in kilograms and the velocities were not in metres per second. Converting this value to a value in joules requires dividing by 1‎ ‎000 to adjust for changing the mass units and by 100 to adjust for changing the velocity units. After dividing by 10, the result is 0.0074184 joules.

The value of Δ𝐾 can be written as an integer value using the unit of ergs. An erg is defined as 1 gram-square centimetre per second squared.1 erg equals 10 joules. This gives the final result of Δ𝑃=74184.ergs

It is important to note that the angle of the sloping plane that the particle moves along is not relevant to determining the changes to the kinetic and potential energy of the system. Had the angle been different, the time at which the speed of the particle had decreased to 29 cm/s would also have been different, but the transfer of energy that occurred in that time would not have differed.

For an open system, energy can be transferred by dissipation due to friction. Such a system is termed a dissipative system. Energy dissipated by such a system equals the sum of the work done on bodies in the system to overcome frictional forces on them.

Energy that is dissipated does not transfer to kinetic or potential energy of the dissipative system, nor to systems that interact with the dissipative system. This implies that the sum of the kinetic and potential energy is not conserved by dissipative systems.

Now let us look at an example where energy is dissipated by frictional force.

Example 4: Calculating Work Done by Friction on a Rough Inclined Plane

A body was projected up a rough inclined plane from its bottom. Its initial kinetic energy was 242 joules. The body continued moving until it reached its maximum height and then slid back down to the bottom. When it reached the bottom, its kinetic energy was 186 joules. Find the work done against friction 𝑊 during the ascent and the gain in gravitational potential energy 𝑃 when the body was at its maximum height.

Answer

Applying the formula 𝐸=𝐾+𝑃+𝑊, we can see that, initially, when 𝑃 is zero and the work done against friction is zero, 242=242+0+0.

And, finally, when 𝑃 is again zero, 242=186+0+𝑊.

The value of 𝑊 is 56. The question asks for the energy dissipated only as the body ascends, however, and 56 joules is dissipated over the entire motion of the body.

Referring to the formula 𝑊=𝐹𝑑𝜃,cos it can be considered that the distance 𝑑 that the body moves while dissipating energy due to friction is the same while ascending as while descending.

The force acting on the body is assumed to act parallel to the direction of the incline of the plane.

It is also assumed that the magnitude of the frictional force 𝐹 is the same while ascending as while descending.

Because of these considerations, the values of 𝑊 for ascent and for descent are equal, so 𝑊 for the ascent is equal to half of the total 𝑊, so 𝑊=2421862=28.joules

If 28 joules of energy is dissipated while ascending, then 𝑃 at the end of the ascent is given by the initial value of 𝐾 less the energy dissipated while ascending: 𝑃=24228=214.joules

In the example, it was assumed that dissipation was equal for motion of the body while both ascending and descending. Assuming that resistive forces responsible for dissipation are of constant magnitude throughout the motion of a body is a simplification, as such resistive forces typically change magnitude in response to the velocity of a body and so are only constant for objects that do not change their kinetic energy as they move.

Let us look at another example where a constant resistive force is assumed.

Example 5: Determining the Velocity of a Body in Motion on an Inclined Plane with a Resistive Force

A car descended 195 m on a slope from rest, which is equivalent to a vertical distance of 14 m. Given that 27 of the potential energy was lost due to resistance and that the resistance remained constant during the car’s motion, determine the car’s velocity after it had traveled the mentioned distance of 195 m. Take 𝑔=9.8/ms.

Answer

It is stated that there is a dissipation of gravitational potential energy, so that only a fraction of the decrease in 𝑃 is transferred to 𝐾. The transfer of energy to 𝐾, Δ𝐾, is given by Δ𝐾=𝑃27𝑃=57𝑃.initialinitialinitial

Using the formula for 𝐾 and for 𝑃, and noting that the initial kinetic energy is zero, it can be seen that 12𝑚𝑣=0+57𝑃12𝑚𝑣=57𝑚𝑔.initial

There is a common factor of 𝑚 on both sides of the equation that can be eliminated to give 12𝑣=57𝑔.

Substituting values from the question for 𝑔 and and rearranging give 𝑣=107×9.8×14.

Taking the square root of both sides gives 𝑣=107×9.8×14=14/.ms

Key Points

  • The work done on a body by a force equals the change in the kinetic energy of the body.
  • The work done on a body by a conservative force is independent of the path traveled by the body.
  • The sum of the work done on a body by a conservative force and the change in the potential energy of the body is zero.
  • The energy of a closed system is equal to the sum of the kinetic energy and potential energy of the bodies in the system.
  • The work done on a body by a non conservative force depends on the path traveled by the body.
  • A dissipative system is one for which some of the work done on bodies in the system does not increase the energy of the bodies.
  • The energy of a dissipative system is equal to the sum of the kinetic energy and potential energy of the bodies in the system and the work done by the system to overcome resistive forces.
  • When the motion of a body is purely due to transfer between potential energy and kinetic energy, the path of the motion of the body cannot be determined from the energy transfer.

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