How Do I Calculate Work Done by Forces in Physics Problems?

[dark_angel]

i don't get it------someone please help

Hi, i think i might have gotten the first aspect of this question right, but the other parts I am 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 what's going on here, they don't 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)


please help

 

[dextercioby]

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.

 

[dark_angel]

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 don't get it, would it be like the same force, but u integrate it positively or something? Bloody hell

 

[whozum]

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]

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 acquire the same "-".

Daniel.

 

[dark_angel]

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 acquire 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 isn't that force going in the same direction as the other one?

omg I am 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

 

[jdstokes]

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.