member 731016
Is that setting the differential work equal to the differential change in kinetic energy? I did not know the work energy theorem could be applied for differentials!erobz said:For some reason, I'm not getting that.
The differential work ##dW## done by the force as it moves a distance ##dx## should be the change in kinetic energy of the pulled rope:
$$ dW = F~dx = \frac{1}{2} \lambda ( x + dx) v^2 - \frac{1}{2} \lambda x v^2 = \frac{1}{2}\lambda v^2 dx $$
$$ \implies F = \frac{1}{2}\lambda v^2$$
?
Many thanks!