Conservation of Energy

The law of conservation of energy states that energy can neither be created nor destroyed; it can only change form. A common example is the apparent “loss” of energy when a ball is dropped from a height. As it falls, it does not bounce back to its original height. So, where has the energy gone? The energy has been transferred and transformed into other forms. As the ball moves through the air, some of its kinetic energy is transferred to the air molecules around it. When it hits the floor, audible sound is created, transferring energy into the air and ground as vibrations. As the ball compresses before bouncing back, the change in its structure results in the emission of heat energy. This example demonstrates how energy is constantly being transferred and transformed all around us.

Use this page to revise the following concepts within conservation of energy:

Types of energy

Energy can exist in many forms, some of the common types of energy include:

Law of conservation of energy

As energy can neither be created nor destroyed, within a closed system, it simply transforms from one form to another.

\[ \sum E_{\text{before}} = \sum E_{\text{after}} \]

This equation represents the law of conservation of energy. It states that the total energy in a closed system remains constant: the sum of all the energy before an event is equal to the sum of all the energy after it. When considering all sources of energy within the system, the total energy does not change, even though it may transform from one form to another.

For example, a ball being dropped from a height can be represented using:

\[\sum{E}_{\text{before}}=\sum{E}_{\text{after}}\]

When the ball is held before it is dropped, all the energy is gravitational potential energy :

\[\sum{E}_{\text{before}}=E_g\]

While the ball is falling, some of the gravitational potential energy transforms into kinetic energy:

\[\sum{E}_{\text{falling}}=E_g + E_k\]

As it hits the ground, all of the relative gravitational potential energy is converted into kinetic energy. Upon impact, the kinetic energy is partially transformed into sound energy and heat energy:

\[\sum{E}_{\text{ground}}=E_k + E_{\text{sound}} + E_{\text{heat}}\]

At any point in this process, the total energy is conserved within the system:

\[\sum{E}_{\text{before}}=\sum{E}_{\text{falling}} = \sum{E}_{\text{ground}}\]


Worked example

A ball is dropped from a height of \(10 \mathrm{~m}\). The ball has a mass of \(0.5 \mathrm{~kg}\). If the speed when the ball makes impact is \(13 \mathrm{~ms}^{-1}\), determine the amount of energy dissipated as sound and heat before impact.

Before the ball is dropped, all the energy is stored as gravitational potential energy:

\[ \begin{aligned} E_g &= mg\Delta h \\ &= (0.5)(9.8)(10) \\ &= 49 \mathrm{~J} \end{aligned} \]

\[ \sum E_{\text{before}} = E_g = 49 \mathrm{~J} \]

The kinetic energy just before the ball makes impact with the ground is:

\[ \begin{aligned} E_k &= \frac{1}{2}mv^2 \\ &= \frac{1}{2}(0.5)(13)^2 \\ &= 42.25 \mathrm{~J} \end{aligned} \]

The law of conservation of energy states that the total energy before falling must equal the total energy just before impact, including any energy dissipated as sound and heat.

\[ \begin{aligned} \sum E_{\text{before}} &= \sum E_{\text{just before impact}} \\ E_g &= E_k + E_{\text{dissipated}} \end{aligned} \]

Therefore:

\[ \begin{aligned} E_{\text{dissipated}} &= E_g - E_k \\ &= 49 - 42.25 \\ &= 6.75 \mathrm{~J} \end{aligned} \]


Hooke’s law

Hooke’s Law describes the force exerted by an ideal spring as being directly proportional to its displacement from the equilibriumposition . It is expressed mathematically as:

\[ F = -k\Delta x \]

Where:

  • \(F\) = force exerted by the spring \((\mathrm{~N})\)
  • \(k\) = spring constant \((\mathrm{~Nm}^{-1})\)
  • \(\Delta x\) = displacement from the equilibrium position \((\mathrm{~m})\)

The negative sign indicates that the force acts in the opposite direction to the displacement. For example, if you compress a spring by pushing it down, it exerts a force upward to try and return to its original length. Similarly, if stretched, the spring pulls back to its equilibrium position.

Assumptions

Hooke’s law only applies to an ideal spring, which assumes:

  1. Linear elasticity: force is directly proportional to displacement within the elastic limit.
  2. The spring is massless.
  3. The spring is frictionless, with no damping losses.

Despite these ideal assumptions, Hooke’s law remains a valuable model for understanding and approximating the behaviour of real-world springs. Many real-life springs obey Hooke’s law over a limited range of displacements.

Force-displacement graph in an ideal spring

Force-displacement graphs provide an alternative way to visualise the relationship between the force applied to an object and its resulting displacement. When assuming an ideal spring , these graphs typically exhibit a linear relationship, with the force directly proportional to the displacement:

Straight line force-displacement graph. Displacement (m) is on the x-axis, and Force (N) is on the y-axis. Line rises up at a 45 degree angle.

As \(F=k\Delta{x}\) (ignoring the negative sign as we are focusing on the force applied to generate the displacement), the spring constant \(k\) can be expressed as:

\[k = \frac{F}{\Delta{x}}\]

This is equivalent to the gradient, or slope, of the graph. By understanding how much force is required to displace a spring, we can determine its spring constant, which measures the spring's stiffness.

The area under the graph is also an indicator of the amount of energy stored in the spring. This is very similar to using area under a curve to calculate to work done:

\[ W = \frac{1}{2}F\Delta x = \frac{1}{2}k\Delta x\Delta x \]

\[ W = \frac{1}{2}k(\Delta x)^2 \]

Straight line force-displacement graph. Displacement (m) is on the x-axis, and Force (N) is on the y-axis. Line rises up at a 45 degree angle. Area between the line and the x-axis is shaded.

Worked example

A spring is compressed by \(0.2 \mathrm{~m}\) from its equilibrium position when a force of \(40 \mathrm{~N}\) is applied to it. Determine the spring constant \(k\) of the spring. Calculate the elastic potential energy stored in the spring when compressed by this amount.

Hooke’s law can be used to determine the spring constant:

\[ k = \frac{F}{\Delta x} \]

Substitute values:

\[ k = \frac{40}{0.2} = 200 \mathrm{~Nm}^{-1} \]

The elastic potential energy can be calculated using the elastic potential energy equation:

\[ \begin{aligned} E_s &= \frac{1}{2}k(\Delta x)^2 \\ &= \frac{1}{2}(200)(0.2)^2 \\ &= 4 \mathrm{~J} \end{aligned} \]


Gravitational potential energy

Gravitational potential energy at or near the Earth’s surface can be calculated simply by:

\[ E_g = mg\Delta h \]

Where:

Far away from the surface of the Earth, the gravitational potential energy of an object, such as the Moon or a satellite , can be written as:

\[ E_g = -\frac{GmM_E}{r} \]

Where:

The negative sign here is a bit weird when you first look at it: negative potential energy is a bit counter intuitive. However, if you pick two distances, one further away and one closer, you’ll see that the potential energy becomes a larger negative number the closer you get. That is, being close to the Earth does indeed give a smaller gravitational potential energy, which is the same relationship as with the equation for gravitational potential energy closer to the Earth.

If you are familiar with Taylor Expansions in calculus, you can think of the expression for gravitational potential energy close to the earth as being the first non-zero term in the Taylor Expansion of the general expression.

This is something we use a lot in physics: when we’re close to something or something’s small, we often get to use a linear equation rather than a more complicated one. This is also why Hooke’s law is such a good approximation for so many springs.