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A block of mass $M$ is taken from point $A$ to point $B$ in a complex path by a force $F$ which is always tangential to the path. We also have coefficient of friction as $K$. What will be the work done by force $F$ when it reaches point $B$ from point $A$? Given that the vertical displacement from $A$ to $B$ is $h$ and the horizontal displacement from $A$ to $B$ is $l$.

In this question, I tried solving the problem using conservation of energy, we know that the total energy will remain constant. So with that, we will have, $$\Delta U_{gravity}+W_{friction}+W_{F} = O$$ But how do you calculate the work done by friction in this case?

Moreover, in the answer, the work done by friction is only dependent on l!!

EDIT: 1.The body is moved very slowly.

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closed as off-topic by John Rennie, Aaron Stevens, JMac, Jon Custer, ACuriousMind Oct 5 at 14:02

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Oct 5 at 14:02
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Friction is not a force from a scalar potential. As such the work done is path dependent so there is not enough information to answer the question.

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  • $\begingroup$ Now there is enough information as the body is moved slowly. $\endgroup$ – Shreyansh Oct 5 at 6:48
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    $\begingroup$ Respectfully, the rate of the bodies movement does not suddenly make friction a conservative force and as such the amount of work done depends on the path taken. $\endgroup$ – Chris Powell Oct 5 at 8:33
  • $\begingroup$ @ChrisPowell In this scenario if the object is moved slowly then the path doesn't matter as long as the path from A to B is monotonically increasing in terms of height. However, this doesn't violate friction being non-conservative. Non-conservative doesn't mean there can't exist multiple paths between two points that take the same amount of work. It just means that you can find two paths with different amounts of work. $\endgroup$ – Aaron Stevens Oct 5 at 11:37
  • $\begingroup$ @AaronStevens Path does matters, because tangential angle (inclination) will change according to x coordinate, thus changing normal force, which is main ingredient in friction force. $\endgroup$ – Agnius Vasiliauskas Oct 5 at 15:53
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    $\begingroup$ The diagram shows a 2D, monotonic path. The post has been edited to say the object moves slowly. An answer can then be obtained here. $\endgroup$ – Aaron Stevens Oct 5 at 22:08
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To calculate the work done by friction, you multiply the friction force times the total length of the path on the surface that the mass is moved. The magnitude of the friction force on a horizontal surface is the product of the normal force on the surface, $mg$, and the coefficient of kinetic friction, which is designated $K$ in the problem, or $F_{f}=Kmg$.

If we assume the surface upon which the mass moves is in the x-y plane, and the the z direction is the vertical displacement, h, then the length of the path for calculating the friction work would be the total distance covered by the mass on the x-y plane.

In the x-y plane a "horizontal" displacement could mean (1) a displacement along the x- axis, (2) a displacement along the y- axis or (3) a combination of both as long as it does not involve a vertical displacement. The problem statement, at least to me, is not clear exactly what the path in the x-y plane is. But based on the answer, the total distance the mass has moved on the on the x-y appears to be I, whatever path that represents.

UPDATE:

Based on the image you just added, the normal force exerted by the mass at any point along the path is $mg$cos θ where θ is the angle between the tangent to the path at any point and the horizontal. This means that only horizontal components (θ= 0) of the displacement along the path contribute to friction work. The vertical components (θ = 90) of the path do not contribute. The end result is that the friction work depends only on the horizontal displacement, $l$.

Hope this helps.

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  • $\begingroup$ I have now added the image.I guess you can now understand it better. $\endgroup$ – Nithish Karthik Oct 4 at 13:37
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    $\begingroup$ @AaronStevens We've been down this path (no pun intended!). But this is the way I feel most comfortable answering. $\endgroup$ – Bob D Oct 4 at 14:09
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    $\begingroup$ No, this is different from last time. Last time the motion was up a straight incline, so we both agreed that $N=mg\cos\theta$. Here that is not true. $\endgroup$ – Aaron Stevens Oct 4 at 14:28
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    $\begingroup$ @BobD Yes, I know. This question is different. The link has motion along straight paths, and the question specifically says to ignore the kinks. In this case we have a smooth path whose direction is changing. Therefore, we cannot say $N=mg\cos\theta$. The correct expression is $N-mg\cos\theta=ma_\text{perp}$, and if the direction is changing $a_\text{perp}\neq0$ $\endgroup$ – Aaron Stevens Oct 4 at 14:56
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    $\begingroup$ @BobD Even then, you still need to assume it's moving extremely slowly, not just friction directly opposing the applied force. Only in the limit where the velocity approaches 0 does $N = mg \cos \theta$. If the path has curvature and there's a velocity, a centripetal force would be required to actually create that motion, finding static equilibrium at each point doesn't work because velocity isn't constant, it changes direction. $\endgroup$ – JMac Oct 4 at 16:02
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Here, at any instant of time, the normal force on the body is $mg\cos\theta$, where $\theta$ is the angle made between the downward direction and the normal force on the wedge direction. Therefore, the work done by friction on moving distance dl is $\mu mg\cos\theta.dl$. Notice, however, that $dl$ could also be written as $\frac{dI}{\cos\theta}$, where $dI$ is the horizontal path length. Hence, the cos term will cancel and on integration will give you your answer irrespective of the path.

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    $\begingroup$ The normal force is not always $mg\cos\theta$ here. That would only be true if the object was moving up a flat incline. If the acceleration has a component perpendicular to the path, then we cannot say $N=mg\cos\theta$ $\endgroup$ – Aaron Stevens Oct 4 at 13:48
  • $\begingroup$ Yes ! θ is not always constant right! $\endgroup$ – Nithish Karthik Oct 4 at 13:55
  • $\begingroup$ @NithishKarthik It's not about constant $\theta$. It is about how acceleration has a component perpendicular to the path, so $N\neq mg\cos\theta$ $\endgroup$ – Aaron Stevens Oct 4 at 14:29
  • $\begingroup$ @NithishKarthik You don't need theta to be constant. The point is that at any instant of time, the cos term will cancel, irrespective of the theta. $\endgroup$ – Rishabh Jain Oct 5 at 14:10
  • $\begingroup$ You also need the object to be moving slowly, which you have not addressed here. $\endgroup$ – Aaron Stevens Oct 5 at 17:48
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The fact that no details on path are given implies that any path should give the same answer. Take a straight line from A to B or move horizontally and then vertically. Either way you get kmgl for the work done by friction. The work done by force F is just Fl + Fh = kmgl + mgh. (This assumes F changes and produces no noticeable acceleration.)

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  • $\begingroup$ Rather than deleting your answer and then posting an edited version, please edit the original answer. You can delete it for editing and then undelete it when you are finished if you like. $\endgroup$ – Chris Oct 4 at 22:08
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In fact the answer is a general result for all particles taken slowly from one point to another when it is on any inclined surface.

The Normal reaction is given as:-$$N=mg\cos\theta$$ This relation is valid as the body is hauled up slowly so the acceleration perpendicular to the surface tends to zero.

$\theta$ is the variable angle made by the the tangent at any point on the surface with the horizontal.$$f=kmg\cos\theta$$

Work done by friction in moving the particle by a distance $ds$ along the path:-$$dW_{friction}=kmg\cos\theta ds$$ Now $$ds\cos\theta=dl$$ So $$dW_{friction}=kmgdl$$ $$W_{friction}=\int kmgdl$$ $$W_{friction}=kmgl$$ We conclude that if the object takes any random path the work done would only depend on horizontal distance between initial and final point.

EDIT 1

As you have asked about the work done by $F$ in the comment section.You can simply use work energy theorem to solve it.

$$W_{total}=\Delta K$$ $$\Delta K=0$$ $$W_{friction}+W_{gravity}+W_{F}=0$$ $$W_{F}=kmgl+mgh=mg(h+kl)$$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Oct 5 at 14:00

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