# Finding Velocity: F=ma or KE = 0.5mv^2

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1. Jun 24, 2017

### Kiwigami

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Here is a made up function of force changing with respect to displacement: F(x) = 2000 - 100x
The force is applied onto an object, pushing it horizontally. Ignore friction and air resistance. The mass is constant. I'm trying to find Velocity with respect to displacement.

Attempt 1:
Knowing F = m*a:
2000 - 100x = m*a
a(x) = (2000-100x)/m
To find velocity, I take the integral of a(x) to get v(x) = [-100(0.5x2 - 20x)]/m

Attempt 2:
Now, I believe that Work = Force * Displacement. W = ∫ from 0 to b of (2000 - 100x)dx = [-50(b-40)b]

Since the object is only sliding horizontally, there's no potential energy, but there is kinetic energy. I set [-100(0.5x2 - 20x)] = 0.5mv^2 and solve for v:

v(x) = sqrt[(2/m) * [-50(x-40)x]]

In the end, I get two different answers. What I would like to know is what am I doing wrong? Did I use illegal math? Did I get the formulas wrong? Did I use the forumlas wrong?

Last edited: Jun 24, 2017
2. Jun 24, 2017

### person123

It seems for the first one you were integrating acceleration with respect to displacement (instead of time)—I don't see how that would tell you the velocity.

3. Jun 24, 2017

### Kiwigami

If I integrate acceleration with respect to displacement, would I get velocity with respect to displacement? Velocity with respect to displacement is what I'm trying to find.

4. Jun 24, 2017

### person123

I'm 90 percent sure that that's false. The connection between acceleration and velocity is that the former multiplied by time specifically equals change in velocity. Therefore, integrating acceleration with respect to time gives you change in velocity. This connection should only work when using time. Note that your first equation seems to fail dimensional analysis (there's the difference of displacement times force and displacement squared, which is meaningless).

I think modelling this situation with a spring might be a creative solution.

Last edited: Jun 24, 2017
5. Jun 24, 2017

### Kiwigami

I see! The units won't work out. In that case, do you think my second attempt is correct?

6. Jun 24, 2017

### person123

I got the same thing both using your second method (which I believe is valid) and imagining it as a spring. (The two methods are actually essentially the same because $PE=½kx^2$ was derived by integrating force over distance). So I'm confident that it's correct, but I would wait until there's someone more experienced to check it if you want to be sure.

7. Jun 24, 2017

### haruspex

No.
$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac 12\frac{d}{dx}v^2$
Integrating wrt x:
$\int a.dx=\frac 12v^2$
Multiplying through by m gives the familiar ∫F.dx=½mv2.