Let’s go back to your garage to make some more physics toys.
You found a nice, rigid spring with a spring constant of 85.4 Newtons per meter.
It is 10 centimeters long when relaxed, and 6 centimeters when fully compressed.
If you compress the spring and put a little 0.01 kilogram army guy on top, how high will
it be launched when you release the spring?
To answer this, we need Newton’s laws, kinematics, and information about springs, so check out
my tutorials on these subjects if you need to.
Then, see what you can get for an answer.
As we said, the spring constant on this spring is 85.4 Newtons per meter.
If the spring is 10 centimeters when relaxed and 6 centimeters when compressed, then releasing
this spring will mean that it expands by a length of four centimeters, or 0.04 meters.
Multiplying this by the spring constant will tell us the amount of force that is imparted
by the spring as it expands this distance.
That will give us 3.4 Newtons as the force imparted on the figurine.
Don’t forget that there’s also gravity pulling downwards, and that’s equal to MG,
so 0.01 kilograms times 9.8 meters per second squared gives us 0.098 Newtons downwards.
That means the net force will still be about 3.3 Newtons upwards.
Now from here, we have to figure out the acceleration that this force will produce.
That’s easy, it’s just F equals MA.
Plugging in 3.3 newtons and 0.01 kilograms, we get three hundred thirty meters per second
squared, a pretty huge acceleration.
But remember, this acceleration only lasts during the time that the figurine is still
in contact with the spring, which is a distance of 0.04 meters, so what velocity will the
figurine have as it leaves the spring, given that it is accelerating at this rate over
this distance?
For that we need kinematics.
Let’s use this equation here, where we can solve for the final velocity, given that the
initial velocity is zero.
We also have the acceleration that we calculated, as well as the distance over which the acceleration
is being generated.
We multiply through, take the square root, and we get 5.14 meters per second as the velocity
of the figurine when it leaves the spring.
Now, with this initial vertical velocity, how high will it go?
Well just as with any question of this nature, there are multiple options available to us,
we just have to use equations where we know everything but one variable, and try to get
some useful information.
As it happens, we could use this same equation again, and we could plug in our initial velocity,
as well as a final velocity of zero, since velocity will be zero at the moment the object
reaches the peak of the trajectory.
We just need to plug in negative 9.8 meters per second squared, making sure to use a negative
value, because the negative acceleration is what will reduce the velocity from this positive
value down to zero, and then we can solve for the distance.
So we square this and bring it to the other side, divide by these two values, and we get
1.35 meters for D. That’s pretty good air for such a little army guy.
Now, this seems like there were a lot of steps to find this answer, but remember, all we
did was find out the force imparted by the spring, the acceleration that force will produce,
the velocity the figurine will attain as a result of that acceleration, and then how
high it will go with that initial velocity with gravity pulling it back down.
As long as we think logically in this manner, the necessary steps to perform any calculation
will reveal themselves.
If you’re still confused, practice, practice, practice.