We all know that $$W = \int_a^b \vec {F} . \vec {ds}$$

We all know that Frictional force = $μN$. Now, finding work done by frictional force(when a body with normal reaction N is moving on surface : $$W = -μN\int_a^b \vec {ds} $$ Now, $\int_a^b \vec {ds} =$ displacement from a to b, which implies that for closed path it is zero and Work done by frictional force is zero and this also implies that it is conservative force.

How is this possible since it is non - conservative force?


2 Answers 2


You replaced the force vector with its magnitude, discarding its direction. The direction is opposite that of the ds vector at all points along the path. So the integrand is not the vector displacement but a scalar.

  • $\begingroup$ Yeah, I forgot that dot product of two quantities is Scalar and I just did dot product but kept the vector. Thanks! $\endgroup$ Feb 13 at 13:31
  • 1
    $\begingroup$ But it's good that you are thinking about these things and testing your understanding. $\endgroup$ Feb 13 at 15:58

Imagine that the motion is around a circle of radius $r$.

You have to evaluate $\displaystyle \int _{\theta_{\rm initial}}^{\theta_{\rm final}} \vec F\cdot d\vec s$.
$\vec F = -\mu\, N \,\hat \theta$ and $d\vec s = r\,d\theta\, \hat \theta \Rightarrow \displaystyle \int _{\theta_{\rm initial}}^{\theta_{\rm final}} \vec F\cdot d\vec s = \int _{\theta_{\rm initial}}^{\theta_{\rm final}}(-\mu\, N \,\hat \theta)\cdot( r\,d\theta\, \hat \theta) = -\mu\, N \,r\int _{\theta_{\rm initial}}^{\theta_{\rm final}}d\theta$.
For one completer revolution the intergal is equal to $2\pi$ and the work done is $-\mu\,N\,2\pi r$ ie minus the magnitude of the force times the circumference of the circle.


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