
#1
May112, 01:32 AM

P: 20

Work done is defined as F vector. dx vector or F dx cos θ where θ is the angle between F vector and dx vector.
But, there is another common formula dU=F.dx Here, dU is the potential energy stored in the object on which the force is acting upon. (Am I correct?) But, the work done on the object, gets stored in it as its internal energy (assuming no heat loss). So. In the case of pushing a block on a rough floor. The force of friction acts opposite to displacement in this case, and so work done would be F(r).dx cos 180 = F(r).dx, i.e. negative. But if we look at it from the energy stored formula, the energy stored in the object would be  (F(r) dx) = F(r)dx., i.e. positive. How come the two energies, although they represent the same thing, have different signs? I am sensing that there is something very wrong with my understanding of things. Help would be appreciated! 



#2
May112, 01:53 AM

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#3
May112, 02:16 AM

P: 973

in fact friction force is non conservative in common language but it is wrong.when sliding friction is acting as in this case friction will do work and the body will come ultimately to rest but from microscopic point of view the atoms or electrons inside the block will get temp. rise and kinetic theory implies that these will be more energetic(kinetic energy) so it may seem that the kinetic energy of block has been destroyed and turned into heat but it is not so.for further reference see feynman lectures vol. 1.




#4
May112, 06:40 AM

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Understanding work done by friction
The equation dU =  F.dx applies only when the force F is a conservative force, like gravity or spring forces. Friction is a nonconservative force, so that equation does not apply here. The work done by friction does not store energy in the object. It adds generally heat energy into the system.




#5
May112, 06:55 AM

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In other words, you can't use that formula for friction. 



#6
May112, 02:38 PM

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#7
May212, 03:26 AM

P: 20

Okay. Right, so friction is non conservative and we can only apply this equation in case of conservative forces.
What happens during walking? Will the work done by friction be negative or postive, or zero? I think it will be zero since there is no actual displacement of the foot. 



#8
May212, 05:06 AM

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#9
May212, 06:26 AM

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One should distinguish between work and a potential of a force.
In the general case work is a functional of a particle's trajectory, which is the solution of the particle's equation of motion, including any kind of forces (conservative, nonconservative or dissipative), i.e., [tex]W[x,C]=\int_C \mathrm{d} \vec{x} \cdot \vec{F}(t,\vec{x},\dot{\vec{x}}) = \int_{t_1}^{t_2} \mathrm{d}t \dot{\vec{x}}(t) \cdot \vec{F}[t,\vec{x}(t),\dot{\vec{x}}(t)].[/tex] As I said, here [itex]\vec{x}(t)[/itex] is a solution of the equation of motion. In the case of a conservative force, the result only depends on the end points of any path in configuration space, leading to a potential for the force, which then is of course only a function of [itex]\vec{x}[/itex], i.e., [tex]\vec{F}=\vec{F}(\vec{x})=\vec{\nabla} V(\vec{x}).[/tex] Here we have [tex]V(\vec{x})=\int_{C(\vec{x}_0,\vec{x})} \mathrm{d} \vec{x}' \cdot \vec{F}(\vec{x}').[/tex] Here [itex]C(\vec{x}_0,\vec{x})[/itex] is an arbitrary path connecting a fixed point [itex]\vec{x}_0[/itex] with the point [itex]\vec{x}[/itex] (of course one has to apply some caution, if one deals with forces defined in regions that are not simply connected). 



#10
May212, 08:39 AM

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There is an alternative view on it: Even in the case of a nonconservative force the force may be the gradient of a potential in a thermodynamical setting.
Consider the idealized situation that the block is thermally insulated from its surrounding and the friction only increases the internal energy U of the block (i.e. that you can neglect the heat capacity of the floor). Then the formula dU=F dx will hold along any differentiable path and the force is the gradient of the potential energy U. However, U will not only be a function of x but in marked difference to classical mechanics also of temperature T. Or, to put it differently, we would have to specify not only the coordinate x of the path but also the temperature T(x) along the path. 


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