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In solution to this question we say $dW/dt=$power that is constant so $F.v$ is constant so with time when velocity increase force must decrease.

But the formula $F.v$ is itself valid in cases of constant forces in case of variable forces it must be written as $dW/dt$

And if we assume force to be constant then work done is $W=F.δx$ then displacement will be of the form (some constant ×t^2) and when we divide both expression with time we find that work is a linear function of time. Which contradicts with answer of the question

So where am I making mistake?

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  • $\begingroup$ $$W=\int\vec F\cdot\vec{\mathrm dx}=\int\vec F\cdot\vec v\,\mathrm dt\qquad\implies\qquad P=\frac{\mathrm dW}{\mathrm dt}=\vec F\cdot\vec v$$i.e. this relation works in general and not just when the forces are constant. You now have a constant power injecting into the system, so you should be able to reason the answer yourself. $\endgroup$
    – naturallyInconsistent
    Commented Dec 8 at 14:44
  • $\begingroup$ @naturallyInconsistent ok but what about my argument in the third paragraph what is my mistake in that $\endgroup$
    – Dev Phalswal
    Commented Dec 8 at 14:49
  • $\begingroup$ If forces are constant, the graph of W v.s. t should be a quadratic curve. The plotted curve is clearly linear, and thus cannot be what you think it is. You are confusing power and work within the same statement. Like, why should you be dividing anything with time if you are concerned about work? You are really saying that power should be linearly increasing with time, and then claiming that it is work. $\endgroup$
    – naturallyInconsistent
    Commented Dec 8 at 14:57
  • $\begingroup$ Oh right yess I just confused power with work thank you $\endgroup$
    – Dev Phalswal
    Commented Dec 8 at 15:03

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