What is the speed of the particle at x=3.8 m?

[astr0]

A force acts on a particle of 1.6kg mass, the force is related to the position of the particle by F=0.5$$x^{3}$$

Find the work done by the force as the particle moves from x=3.8 m to x=2.0 m.

By plugging each value of x into the force equation I get that:
F at 3.8 m = 27.436 N
F at 2.0 m = 4.000 N

I don't know where to go from here.
I have tried using the average of those two forces, and that was not correct.

 

[rock.freak667]

How do you define work done in terms of an integral?

 

[astr0]

Work = $$\int F*dx$$
Which gives $$\frac{1}{2}F^{2}$$
But how does that help me?

 

[rock.freak667]

Work = $$\int F*dx$$
Which gives $$\frac{1}{2}F^{2}$$
But how does that help me?

So you know that

$$W= \int F dx.$$

You know F in terms of some function x and they told you the limits for the x values.

 

[fisixC]

Simplified, W = integral(F dx) with the limits of integration. Therefore your F = 0.5(x^3) represents the function being integrated as with the change in x which is given for you. Thus, when you find the definite integral with the bounds given as with the function F, you have your work.

 

[astr0]

I understand now. I wasn't making the connection. Thanks.

 

[Matterwave]

Work = $$\int F*dx$$
Which gives $$\frac{1}{2}F^{2}$$
But how does that help me?

Er that integration is done incorrectly. Integrating F with respect to dx will not give you .5F^2. You are incorrectly applying the power rule for integration...

 

[astr0]

I realize that now.

 

[astr0]

The second part of the problem states At x= 2.0 m the force points opposite the direction of the particle's velocity (speed is 12 m/s). What is its speed at 3.8 m?

If the force points opposite, shouldn't the particle be slowing down?