Suppose we have a block of mass $M$ and we are moving it up a curve, very slowly ($a=0$). The surface is not smooth, and coefficient of friction is $\mu=\mu_s=\mu_k$.

To move the block we apply a force $F$ to the block tangential to the surface(on which the block is, at that time). If we take the angle of slope to be $\theta$ (of the surface), then the forces acting are friction$f$, gravity/weight $Mg$ and external force $F$.


This is where the problem comes. If we want to find out the work done by friction, then $$W_{friction}=\int\vec{F_{friction}}.\vec{dr}=-\mu Mg\int dr cos\theta=-\mu Mg\int dx$$ But when I use the work-energy theorem, then $$W_{friction}+W_{gravity}+W_{ext}=0$$ $$W_{friction}=-W_{gravity}-W_{ext}$$ $$W_{friction}=Mgh-F\int dr$$ When I get the work done by friction using integration then it tells me it doesn't depend on the type of path and path length, but only horizontal $x$ covered. When I use the work-energy theorem, it tells me that it depends on height $h$ till which the block goes and path length $r$. What is going wrong? Could anyone hint me?

Thanks in advance.

  • $\begingroup$ How you calculate $-F\int dr$? $\endgroup$
    – ABC
    Commented Mar 27, 2013 at 8:49
  • $\begingroup$ @exploringnet: that part- force $F$ is constant in magnitude and it always pushes the block in the direction of the slope by an amount $dr$. After that $dr$ distance the angle of slope changes and the direction of $F$ changes and the process continues, and hence $\int dr$ amounts to the total path length. In other words $\int dr$ adds up to curve length just as small dots add up to a figure. $\endgroup$ Commented Mar 27, 2013 at 8:53

1 Answer 1


Let at any angle the slope of path is $\phi$ :

So, $$-F\int dr =-\left[\int F\cdot\text dr\cos\phi +\int F\cdot\text dr\sin\phi\right];$$

Now as $\Delta K.E.=0;$ (moving slowly);

Net work by horizontal forces = Net work done by vertical forces =0;

Now , you can see vertical work = $\int F\sin\phi\cdot \text dr - Mgh =0$

and horizontal work =$\int F\cos\phi\cdot \text dr $ must be cancelled out by the frictional work , which shows work of friction just depends on horizontal distance. So,The first method was right and second one was not completed.

  • $\begingroup$ do the terms vertical work and horizontal work refer to work done by vertical components and horizontal components of all the forces? The idea seems ambiguous because I'd heard work was scalar, and more than that, the work kinetic energy theorem should hold for forces, not components. I mean we are taught work done by all forces = change in KE and not that, work done by all vertical forces = change in vertical KE (using $V_y$ instead of $v$) and same for x. Are you sure this holds? $\endgroup$ Commented Mar 27, 2013 at 9:08
  • $\begingroup$ Yes . I mean't that only .I'll edit it. I just took works of horizontal forces and vertical forces separately as they can be tackled easily. $\endgroup$
    – ABC
    Commented Mar 27, 2013 at 9:12
  • $\begingroup$ please use $\vec{dr}=[drcos\phi ,drsin\phi]$, this should be used. otherwise $\int dxcos\phi$ causes confusion. $\endgroup$ Commented Mar 27, 2013 at 9:15
  • $\begingroup$ Yes the work for sure can be broken down into two mutually perpendicular directions . As let $F_x \ and\ F_y $are forces acting which brign about a change in v from 0 to $V_x,V_y$ So $$F_x \times x + F_y \times y = 1/2 m \sqrt{{V_x}^2 + {V_y}^2}$$. That can be written again as $F_x \times x == 1/2 m{V_x}^2$ and $ F_y \times y = 1/2 m{V_y}^2$ $\endgroup$
    – ABC
    Commented Mar 27, 2013 at 9:21
  • 1
    $\begingroup$ And Sorry that I have put the \sqrt symbol on the net velocity expression in the K.E.[it get squared :p] .It is $$F_x \times x + F_y \times y = 1/2 m({V_x}^2 + {V_y}^2)$$. $\endgroup$
    – ABC
    Commented Mar 27, 2013 at 9:29

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