Force Fx exerted on a particle

1. Jul 24, 2013

bestchemist

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

The figure shows the force Fx exerted on a particle that moves along the x-axis. Draw a graph of the particle's potential energy as a function of position x from 0m to 1.1m. Let U be zero at x=0m.

2. Relevant equations

W = F*d
U = mgh

2. Jul 24, 2013

PhanthomJay

Attempt please. Your relevant equations are not correct when the force is not constant. The force is not mg. Are you familiar with calculus or other means of calculating work or PE for non constant (assumed conservative) forces?

3. Jul 24, 2013

bestchemist

4. Jul 24, 2013

voko

It is given, via the diagram.

5. Jul 24, 2013

bestchemist

since the slope from 0 to 0.5 is 4 then the force is F = 4x and the integral of that is 2x2 right?
from 0.5 to 1 the force is what? since the slope there is -4, will it be F=-4x?

6. Jul 24, 2013

PhanthomJay

yes, it is actually 2x^2 + k, where k is a constant, and since U =0 when x is 0, then k is 0, so U = 2x^2 between x = 0 and x = 0.5, an upward facing parabola
Not quite, are you familiar with the equation of a straight line , y = mx + b, where m is the slope of the line and b is the value of y at x = 0 (the y-intercept of the extended line)?

7. Jul 25, 2013

bestchemist

Soooo...... From 0.5 to 1, F = -4x +4 since the force at 0.5 is 2 so the y intercept is 4!
And then the integral of F will be U=-(2x^2+4x +k) right??
But somehow when I graph this U graph... It doesn't look right. When I plug 0.6 in for x I got 3.12

8. Jul 25, 2013

voko

When you have $$f(x) = \begin{cases} a(x), 0 \le x < 0.5 \\ b(x), 0.5 \le x < 1 \\ c(x) \end{cases}$$ then what is $$\int_0^x f(x) dx$$?

9. Jul 25, 2013

PhanthomJay

yes
you mean U = -2x^2 +4x + k, I think, where k can be found using the knowledge gained in the first part, U = 0.5 when x = 0.5, solve for k.
Try again, using the corrected equation for U.

Don't forget you then still need to graph the PE from x =1 to x = 1.1.

As a check of your calculus, remember that the area under the force-displacement curve from F=0 to F=x, is the PE at x.

I'm not so good at calculus, either, just the basics. Haven't used it much since college days.

10. Jul 27, 2013

bestchemist

Thank you guys,

I think I got it!

but for x=1 to x=1.1
isn't the force is zero so the PE is zero too? the slope is zero so F=0 integral of that will be zero too isn't it?

11. Jul 27, 2013

voko

Zero derivative does not mean that the function is zero. You want your potential energy to be a smooth function, so it should not be zero at x > 1 because it was not zero at x < 1.

12. Jul 27, 2013

bestchemist

Sooo
∫F= 0
so U = 0x +k
so U =1?

13. Jul 27, 2013

voko

Can you answer the question in #8?

14. Jul 27, 2013

bestchemist

nope :(
I got a(x) and b(x) but I don't know c(x)

15. Jul 27, 2013

voko

c(x) = 0, as given.

16. Jul 27, 2013

bestchemist

U = ∫11.1 0
U= 0x + k
isn't it?

17. Jul 27, 2013

voko

$U = \int_{1}^{1.1} 0 dx$ has almost nothing to do with $U = \int_0^x f(x) dx$ as requested in #8. It cannot be the answer.

18. Jul 27, 2013

bestchemist

I don't know what could it be then
I feel discourage :(
Can you tell me the answer and how you get it step by step, please...........

19. Jul 27, 2013

voko

You need to find this: $$U(x) = - \int_0^x f(x) dx$$ The complication is that $f(x)$ is not defined as a simple formula:
$$f(x) = \begin{cases} a(x), 0 \le x < 0.5 \\ b(x), 0.5 \le x < 1 \\ c(x) \end{cases}$$

Obviously, when $0 \le x < 0.5$, you can say $U(x) = - \int_0^x a(x) dx$.

But when $x > 0.5$, you cannot say that $U(x) = \int_0^x a(x) dx$, nor can you say that $U(x) = \int_0^x b(x) dx$. However, for any $f(x)$ you can say $\int_0^x f(x) dx = \int_0^{0.5} f(x) dx + \int_{0.5}^x f(x) dx$, so, for $0.5 \le x < 1, \ U(x) = - \int_0^{0.5} a(x) dx - \int_{0.5}^x b(x) dx$.

Now you have to find $U(x)$ for $x \ge 1$.

20. Jul 27, 2013

bestchemist

so for x>1
U(x) = -∫0 0.5 ax dx - ∫0.5 1 bx dx - ∫1x cx dx

right?