# Mechanics 6: Work and Energy

How can we determine if a force does work? What is the definition of work? The definition of work is when a force is exerted on an object over some distance. So work can be thought of as (force \(\times\) distance), and it has units of N \(\cdot\) m or newton-meters. So with this definition, we can calculate the amount of work done by a force. So another question we can ask is, “can work be stored?” That way, we could keep a box of it ready when we need work done. This “stored work” is called *energy.* It also has units of newton-meters. Since this is such a common thing in Physics, we define a new unit, called the joule (pronounced “jewel”). Since it is named after the physicist Robert Prescott Joule, we abbreviate it with a capital letter, J. So as you may have guessed, a newton-meter equals a joule.

Energy exists in two basic forms: kinetic and potential. Kinetic means “motion,” potential means “could move.” Kinetic energy is the energy of motion, and is defined as \(\frac12 m v^2.\) Where does this come from? The definition of work/energy is \(E=Fd,\) and from our equations of motion, for a body that has a force applied and starts from rest (zero velocity), we have \(d = \frac 1 2 a t^2.\) Combining the two, we get

$$E = Fd = F \bigg(\frac 1 2 a t^2\bigg)$$

and we know from Newton’s law that \(F = ma,\) so we have

$$E = (ma) \bigg(\frac 1 2 a t^2\bigg)$$

Rearranging

$$E = \frac 1 2 m (a t)^2$$

And also remember from the equations of motion, \(v = at.\) Substituting gives

$$E = \frac 1 2 m v ^2$$

The other type of energy is Potential Energy, which can be thought of as “stored” energy. It can be stored in many ways — electrical, chemical, gravitational, inertial, etc. Here we will talk about gravitational potential energy. Start with the first equation above

$$E = Fd = mad$$

If the energy is stored as gravitational potential, then it is due to an object being lifted (against the force of gravity) to some height. So, here the acceleration is \(g,\) and the distance is \(h,\) the height lifted. Substituting these in gives the most common way of expressing potential energy,

$$E = mgh$$

Some books use \(K\) for kinetic energy and \(U\) for potential energy, so you may see

$$K = \frac 1 2 m v^2 \qquad {\rm and } \qquad U = mgh$$

### Work-Energy Theorem

When work is done on an object, that work (energy) has to come from somewhere. The process converts kinetic or potential energy into work. So if we measure the kinetic energy before and after, then the difference is the amount of work done. To give it as an equation

$$\Delta K = K_f – K_i = W$$

That is, the work done is equal to the change in kinetic energy. The change in kinetic energy is calculated as (final – initial).

### Conservation of Energy

The Law of Conservation of Energy is, “Energy cannot be created nor destroyed” — but it can change forms. Kinetic energy can be converted to potential energy and vice versa. For example, as a car drives up a hill, its kinetic energy is converted into potential energy.

### Examples

Let’s look at a free-fall problem. We’ll do it with the equations of motion, and then again with conservation of energy and compare. Say we’re up on a ladder, a height \(h\) above ground, and we drop a hammer. What is its velocity when it hits ground?

First, with the equations of motion. Acceleration is only due to gravity, so \(a=-g\) (downward). The hammer is dropped, so initial velocity is \(v_0 = 0.\) Initial position is \(y_0 = h,\) so we have

$$v(t) = -gt \qquad {\rm and} \qquad y(t) = -\frac 1 2 g t^2 + h$$

So the time needed to reach the ground \((y=0)\) is calculated with \(\frac 1 2 g t^2 = h,\) so \(gt^2 = 2h,\) and then \(t=\sqrt{2h/g}.\) So the velocity at this time is

$$v(t) = -g \sqrt{\frac{2h}g}$$

we can take the \(g\) under the radical as \(g^2,\) giving

$$v(t) = – \sqrt{ \bigg(\frac{2h}g\bigg) g^2} = – \sqrt{2gh}$$

The negative sign means that it’s moving downwards.

Now let’s work it again with the principle of conservation of energy. Our hammer starts at a distance $h$ above ground, so its potential energy is \(mgh,\)

and since it’s not moving, its kinetic energy is zero. When it reaches ground, this potential energy is converted into kinetic energy, \(\frac 1 2 m v^2.\) Setting these equal, we get

$$\frac 1 2 m v^2 = mgh$$

Notice that the \(m\) terms cancel. This tells us that acceleration due to gravity does not depend on mass; i.e., all masses accelerate the same. Galileo discovered this long ago and said that in a vacuum, a feather and a hammer would fall at the same rates. Astronaut David Scott did that experiment on the moon during Apollo 15, and showed that Galileo was right. And, of course, you can see it on YouTube. Solving for \(v\) we get \(v^2 = 2gh,\) or

$$v=\sqrt{2gh}$$

Since energy is a scalar, there is no direction, so we have to insert the direction at the end of the calculation, like so

$$v=-\sqrt{2gh}$$

as before.

#### (With Numbers)

Say that our hammer starts from 5 m, then we get \(v = -\sqrt{2g(5)} = -9.9\,\)m/s.

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The actual pages are quite good. Clear, concise, the page layout is easy to read.

the headers could use some design work. I know that doesn’t seem important to the work, but I would consider finding an artist to redo them.

Point well taken! I need to see if I can find such an artist….