Hi, i think i might have gotten the first aspect of this question right, but the other parts im not too sure about...ie i have no idea what its going on about.

Question: An object is attracted toward the origin with a force given by F_x = -k/(x^2)

part a: Calculate the work done by a force F_x when the object moves in the x-diection from x_1 to x_2.

for this i found the integral of the force with limits x_1 to x_2 , i ended up with something like

work = k( (1/x_2) - (1/x_1) )

this might be totally wrong though

partb: if x_2 is greater than x_1, is the work done by F_x positive or negative?

i got negative, because it turns out negative for the integral answer.

partc: The only other force acting on the object is a force that you exert with your hand to move the obect slowly from x_1 to x_2. How much work do you do?

i have absolutely no idea whats going on here, they dont tell u what this force is???

partd: if x_2 is greater than x_1 is the work you do positive or negative? (in relation to part c)

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dextercioby
Homework Helper
a) is correct.So is b).For point c),what sense has the force you exert wrt the sense of the force in the points b) & a)...?

If you figure that out & count the fact that the problem does not mention other force,nor the kinematics of the body,u can solve c) & d) easily;

Daniel.

dextercioby said:
a) is correct.So is b).For point c),what sense has the force you exert wrt the sense of the force in the points b) & a)...?

If you figure that out & count the fact that the problem does not mention other force,nor the kinematics of the body,u can solve c) & d) easily;

Daniel.

thanks for that verification, i'm glad i got something right.

arent u doing like negative work, since your taking energy away....or is it positive? If so, i still dont get it, would it be like the same force, but u integrate it positively or something? Bloody hell :grumpy:

Work is the integral of F with respect to distance. Here your distance is x_2-x_1, and the force F is -k/x^2. Integrate F with respect to x and evaluate at the points x_2 and x_1. Going WITH a force increases potential, going AGAINST a force decreases potential.

dextercioby
Homework Helper
There's no convention here,like in thermodynamics.You just gave to apply calculus & the definition.

And yes,it is the same force in absolute value,but it has a negative sign (compared to the I-st),therefore the work done would aquire the same "-".

Daniel.

dextercioby said:
There's no convention here,like in thermodynamics.You just gave to apply calculus & the definition.

And yes,it is the same force in absolute value,but it has a negative sign (compared to the I-st),therefore the work done would aquire the same "-".

Daniel.

what the?!?!? but how can it be the same magnitude of force, i mean the question dosent specify anything about the force in terms of its magnitude, how can we deduce that it is infact the same magnitude with a negative?

and isnt that force going in the same direction as the other one?

omg im getting even more confused.

so what youre saying is that its the same force with magnitude Fx=k/(x^2)
and so i do the same integration thing.

i end up with k(1/x_1 - 1/x_2)

and the next part is positive work

Hi dark_angel,

The fact that the object moves slowly'' from $x_1$ to $x_2$ suggests that it is not accelerating, and therefore the force from your hand balances the attractive force. The work done by your hand is thus
$\int_{x_1}^{x_2}\frac{k}{x^2}\thinspace\mathrm{d}x = k\left(\frac{1}{x_1} -\frac{1}{x_2}\right)$
as you said.