Mariah Sparks

2022-07-23

Why is small work done always taken as $dW=F\cdot dx$ and not $dW=x\cdot dF$ ?

Steven Bates

You can see this from the second Newton's law:
$m\stackrel{¨}{\mathbf{x}}=\mathbf{F}\left(\mathbf{x}\right)$
Now I would like to integrate this equation of motion with respect to time, to arrive at the energy conservation. To do so I multiply both sides with $\stackrel{˙}{\mathbf{x}}$:
$m\stackrel{¨}{\mathbf{x}}\cdot \stackrel{˙}{\mathbf{x}}=\mathbf{F}\left(\mathbf{x}\right)\cdot \stackrel{˙}{\mathbf{x}}$
and finally integrate:
$m\int dt\stackrel{¨}{\mathbf{x}}\cdot \stackrel{˙}{\mathbf{x}}=\int dt\mathbf{F}\left(\mathbf{x}\right)\cdot \stackrel{˙}{\mathbf{x}}$
The l.h.s. gives me the kinetic energy. The r.h.s. gives me exactly the integral in question:
$\frac{1}{2}m{\stackrel{˙}{\mathbf{x}}}^{2}=\int d\mathbf{x}\cdot \mathbf{F}\left(\mathbf{x}\right)$So the work done by the force is the kinetic energy of the particle (up to an integration constant representing its total energy).

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