# Homework Help: Force and integration

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1. Feb 18, 2017

### Elias Waranoi

Hello! First time poster, please treat me well! I've already solved the problem below on my second attempt with the help of kinetic energy but I want to know why my first attempt gives a wrong answer.

1. The problem statement, all variables and given/known data

A force in the +x-direction with magnitude F(x) = 18.0 N - (0.530 N/m)x is applied to a 6.00-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box. If the box is initially at rest at x = 0, what is its speed after it has traveled 14.0 m?

2. Relevant equations
a = F/m

3. The attempt at a solution
F(x) = 18.0 - 0.530x =>
a(x) = 18.0/m - 0.530x/m =>
Integrate =>
v(x) = 18.0x/m - 0.530x2/2m = 3x - 0.530/12 x2
v(14) = 33.3

Why is this incorrect? I'm guessing that the integral of a(t) is velocity but the integral of a(x) is something else.

Last edited: Feb 18, 2017
2. Feb 18, 2017

### Staff: Mentor

Welcome to the PF

When you integrate x, you don't just get x^2...

3. Feb 18, 2017

### Elias Waranoi

Oops, I edited my question. The result is still wrong though.

4. Feb 18, 2017

### Staff: Mentor

Looks like it's just a math error when you plug in x=14. Try again?

5. Feb 18, 2017

### Elias Waranoi

I typed this into wolfram alpha "3*14 - 0.53/12*14^2" and got 33.3. When I use W = K2 - K1, W = F*s with the average force from x = 0 to x = 14 I get the velocity at 14 meters to be 8.17 which is the correct answer.

6. Feb 18, 2017

### Staff: Mentor

Exactly. If you integrate F(x)dx, you can calculate the work and thus the change in KE.

7. Feb 18, 2017

### Staff: Mentor

You should use parenthesis, I believe. x^2 is not in the denominator...

8. Feb 18, 2017

### Staff: Mentor

Ah, so the choice of integral is wrong, regardless of errors in the integration. Thanks Doc.

9. Feb 18, 2017

### Elias Waranoi

Ahh, so if F(x)dx integrated is work I'm guessing that integrating a unit1 with unit2 equals unit1*unit2. So when I integrated a(x)dx I actually got meter2/time2 instead of velocity. Is this correct?

10. Feb 18, 2017

### Staff: Mentor

That's true.

11. Feb 18, 2017

### haruspex

It is easy to get confused here.
It depends whether you are thinking of a as the physical entity acceleration or as purely a mathematical function.
As a physical entity, it's not whether you think of a as a function of time or of displacement that matters. What matters is what you integrate with respect to, as you figured out. Integration wrt a variable is akin to multiplication by that variable. ∫a.dt gives Δv, while ∫a.dx gives Δ(v2)/2.
If you are given a as a function of displacement, ∫a.dt still yields velocity, but how are you to perform the integral? E.g. if told a=kx3, your integral is ∫kx3.dt. Since you do not know what x is as a function of t this is not directly solvable. If you are told the total displacement then you can use ∫a.dx instead to find the velocity, but what if you are only told the elapsed time? In that case you would have to write out and solve the differential equation.