The connection between energy and force
Our operational definition of energy involved the ability of an object to change itself, or change the environment in some way.
Does a force always, or only sometime result in a change in the environment?
"Work"
When the point of application of the force moves, then there is a change in the environment. So, let's consider the quantity: $$F\cdot d \equiv W$$ where $F$ is the force acting on an object and $d$ is the distance that the "point of application" moves. We call their product work '$W$'. Does this behave like you'd expect energy to behave?
Let's say you want to lift one or more books from a table to a shelf one meter above the table. If you double the number of books you lift, would you think the amount of energy needed to lift the books would be...
- double,
- the same, or
- half as much?
And look at what happens to the "work".
Units of work
Work is defined as: $$\text{Work}=F\cdot d$$ The units should be Newtons times meters.
This unit (for energy) is used so often, that it gets its own name:
$$1 \text{ Newton}\cdot\text{meter}\equiv 1 \text{ "Joule"}=1\text{ J}.$$
"Weight"
The force that you measured with the spring scales was the force due to gravity acting on however much mass you had.
This force gets a special name: It is the object's weight, $w$.
An object's weight is a force. It is the force (as measured by a spring balance) that gravity exerts on the object.
You measured
- $w$ -- the spring force in "Newtons" due to some...
- $m$ - mass in kilograms.
- You found the slope, $k$, of the resulting line:
$$w=k\cdot m$$
where the constant, k, was (to one decimal place) 9.8. Our constant, $k$, has to have units of N/kg to make this equation come out right.
It looks like the same number as the acceleration of all objects (near Earth's surface), $$g=9.8 \text{ m/sec/sec},$$ due to gravity on Earth. Hmmm. This would only be true if $$\frac{\text{N}}{\text{kg}}=\frac{\text{m}}{\text{sec}^2}$$ which means that $$\text{1 N}= 1 \frac{\text{ kg}\cdot \text{m}}{\text{sec}^2}.$$
Let's run with this for the time being.
This means that the force of gravity on an object could be written as
$$\text{weight}\equiv w = mg$$
What does that mean for Gravitational Energy which we said was equal to the force of gravity times height
$$\text{GravE}=wh?$$
This would have to mean:
$$\text{GravE}=mgh.$$
You should verify mentally that this behaves like we'd expect the destructive capacity of books on shelves to behave.
How much work does it take to lift something?
The work, $F\cdot d$, that it takes to lift an object is the force of gravity on it--its weight--times the distance moved, which is the height, $h$, that we're lifting it through. Oh, this is just our expression for GravE: $$wh=(mg)h.$$
On the moon, an object that weighs 6 lbs on Earth weighs (according to a spring scale) only 1 lb. Approximately what is the acceleration of freely falling objects on the moon?
Video of astronauts dropping things. There is, by the way, no air on the moon.