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## High school physics - NGSS

### Course: High school physics - NGSS > Unit 4

Lesson 3: Predictions using energy# Calculating velocity using energy

Learn how you can calculate the launch velocity of an object by using the total energy of a system. Energy that is conserved can be transferred within a system from one object to another changing the characteristics of each object, like velocity.
Created by Sal Khan.

## Want to join the conversation?

- Is this called a certain equation or not? Is it applicable to every situation? What is Δx equal to? (in most situations; in here it means spring compression)(2 votes)
- Δx or dx stands for 'delta x' which indicates displacement. This is calculated from taking the initial x position to the final x position (final x - initial x). In this case for the spring, the initial position is the uncompressed mark and the final position is the compressed stage, so delta x would measure how much the spring moved (which in this case more displacement = more compression).(2 votes)

- why is it kΔx^2/m - 2gh2 at5:55(2 votes)

## Video transcript

- [Instructor] So we have a spring here that has a spring constant
of four Newtons per meter. What we then do is take a 10-gram mass and we put it on top of the spring and we push down to compress
the spring by 10 centimeters. We then let go. And what I'm curious about
is what is going to be the magnitude of the
velocity of our ball here, of our 10-gram mass, right as the spring is no
longer compressed or stretched, or essentially when the
ball is being launched? Pause this video and see
if you can figure that out. And I'll give you a hint, the
energy in this first state, the total energy has got to
be equal to the total energy of this second state. We can't create or destroy energy. All right, now let's work
through this together. So let's call this first
scenario state one. So in state one, what is the
total energy going to be? Well, it's going to be the sum of the gravitational potential energy, so that's mg times the
height in state one, plus our elastic potential energy, that's 1/2 times the spring constant times how much we've compressed that
spring in state one squared, plus our kinetic energy, so that's 1/2 times our
mass times the magnitude of our velocity in state one squared. And that has got to be equal to, as we just talked about, the
total energy in state two. Well, what's that going to be? Well, that's your
gravitational potential energy in state two plus your elastic
potential energy in state two plus your kinetic energy in state two. Now let's think about which
of these variables we know and which ones we need to solve for. So, first of all, mass,
your spring constant, the strength of your gravitational field, well, we know what these are going to be. These are going to be our
mass is equal to 10 grams. The strength of the gravitational field, also the acceleration due to gravity near the surface of the earth, is
9.8 meters per second squared. Our spring constant is
four Newtons per meter, and I like to remind
myself what a Newton is, a Newton is kilogram
meter per second squared. So this is also equal to four kilogram meter per second squared, and then we also have a meter over there. And actually, these
meters will cancel out. And that's useful
because it's reminding us that we want everything to
be in kilograms and meters. And with that in mind, actually, let me rewrite our mass right
over here as 0.01 kilograms. And then let's think about
what's going on specifically in each of those states. So what is going to be our
h one, our initial height? Well, I didn't give it to you, but what really matters is the difference between h one and h two. So we could just define
h one right over here to be equal to zero. So let me write that down. H one is equal to zero. And if we say that, then
what is h two going to be? H two would then be
equal to 10 centimeters, but remember, we want everything
in kilograms and meters, so 10 centimeters is the
same thing is 0.1 meters. What is our spring compression in scenario one going to be? Well, that is going to be 10 centimeters. But once again, we wanna
write that in terms of meters. So I'll write that as 0.1 meters. And then what is our spring compression in scenario two going to be? Well, we're completely
uncompressed and unstretched, so that is going to be zero meters. And then what is going to be our velocity, or at least the magnitude of
our velocity in state one? Well, we're stationary, so
it's zero meters per second. And what is going to be the magnitude of our velocity in state two? Well, that's exactly
what we wanna solve for. That is our launch velocity. So let's see if we can simplify this and then solve for v two. So we know that h one is equal to zero. So that simplifies that right
over there, that term is zero. We know that v one is zero. So that simplifies that term there. We know that delta x in
scenario two is equal to zero. So that simplifies that
term right over there. And so we can now rewrite all of this, and I'll switch to one color
just to speed things up a little bit, as 1/2 k
times delta x one squared is equal to mgh sub two
plus 1/2 mv sub two squared. And now let's just try to
solve for this character. So let's subtract mgh
sub two from both sides. So we're going to have 1/2 k
times delta x sub one squared minus mgh sub two is equal
to 1/2 mv sub two squared. Now let's see, if we multiply
both sides by two over m, then that will get rid
of this 1/2 m over here. So let me multiply that. I'm kind of doing two
steps at the same time. One way to think about it
is I'm multiplying times the reciprocal of the
coefficient on the v squared. So right over here, it's m over two, so the reciprocal is two over m. I'm gonna have to make sure
that I'm gonna multiply it times that whole side. And so what do we now have? This is going to be equal to, let's see, it's going to be k delta
x sub one squared over m minus two gh sub two is going to be equal to v sub two squared. And then to solve for our launch velocity, we just take the principal route, the square root of both sides. So I could say v sub two, which is equal to our launch velocity, is equal to the square
root of our spring constant times delta x sub one squared
over our mass minus two times the gravitational field times h two. And so we just have to
plug in the numbers now. This is going to be equal to, and I'll switch colors
just to ease the monotony. I'm gonna use a new color right over here. This is going to be equal
to the square root of, and I wanna make sure
all the units work out. So I'm actually gonna write this version of my spring constant, so I
can work with all the units. So it's gonna be four
kilograms per second squared. And now delta x one we know is 0.1 meters. So if we square it, it's going to be times
0.01 meters squared. And then all of that is
going to be over our mass, which we know is 0.01
kilograms, 0.01 kilograms, and then minus two times 9.8
meters per second squared times height in the second scenario, which we already know is 0.1
meters, so times 0.1 meters. Let me extend this
radical right over here. And let's look at the units first to just to make sure we're
getting the right units. So this kilogram is going to
cancel with this kilogram. And then we have, over here, we're going to have
something that's in terms of meters squared per second squared, and then over here, we're gonna have something in terms of meters squared per second squared. So that makes sense. We're gonna have a difference of two meters squared per second squared. And then when you take
the principal route, you're going to get meters per second, which is the unit for the
magnitude of velocity. So now we just have to
get our calculator out and calculate this. So the .01s will cancel out. So this part right over
here is going to be four. And then from that, I am going to subtract
two times 9.8 times .1, close the parentheses. That's going to be equal to that. And then I needed to take the square root of all of that business, and
I get this right over here. So this is approximately
1.43 meters per second, 1.43 meters per second. And we're done.