Couldn't think of a great way to frame my question, but I'm having trouble understanding why non-conservative work is a negative value.

For this problem I'm working on, there's a box moving at speed $v_i$, toward the right, on a surface with friction. I'm trying to figure out how far the box will travel until $v_f$=0.

Assuming gravity, the normal force, and friction are the only forces acting on the box, I know:

$N$ - $F_g$ = 0 -->$N$ = $F_g$ = $mg$

W(nonconservative) = $F$ $x$ cos($\theta$) = $F_{friction}$ $x$ cos($\theta$)

Setting the right as my positive direction, I assume that the force due to friction is therefore pointing left, in my negative direction. But the box is still moving right because of $v_i$ Therefore since my force vector and my direction vector are pointing in opposite directions, cos($\theta$) = cos(180$^o$).
Therefore, I'm getting $W_{nc}$= -$F_f$ $x$ cos(180$^o$) = $F_f$ $x$

However, I'm told that $W_{nc}$=-$F_f$ $x$

                           **^^Where is this negative sign coming from??**

Thanks for all your help. Sincerely, First time physics student trudging slowly through a semester of Physics I

  • $\begingroup$ I think, you should ask yourself, which work you are trying to compute. Is it the work done by the box on its surroundings? Or is it the work done by the friction force on the box? The two works have equal absolute value but opposite sign. $\endgroup$
    – Photon
    Mar 30 '17 at 8:56

Work done $W$ is related to the force $\vec F$ and the displacement $\vec x$.
$W=\vec F\cdot \vec x$

What you are then doing is introducing a negative sign twice.

You either say that $W=Fx\cos 180^\circ =-Fx$ which is a direct use of the definition of work done or $W=(-F\hat x)\cdot(+x \hat x)$ using the fact that the frictional force is in the negative direction $(-\hat x)$ and the displacement is in the positive direction $(+\hat x)$ but not both together.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.