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My question is this, $P= dW/dt.$

dW=F.dr

That leads to P= F.v But how?!

The book (Schaum's theoretical mechanics) stated that this formula is true. But I think it is true just in case that F is constant, otherwise we have to apply the rule of the product of derivatives for dot product!

Can any one help me and illustrate the bug picture for me? enter image description hereenter image description here

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    $\begingroup$ $F$ is assumed constant during each slice of time ${\rm d}t$. $\endgroup$ – John Alexiou Apr 29 at 0:28
  • $\begingroup$ But can we apply this formul for variable force fields? $\endgroup$ – Sohaib Ali Alburihy Apr 29 at 0:37
  • $\begingroup$ Sorry I meant formula* not formul $\endgroup$ – Sohaib Ali Alburihy Apr 29 at 0:50
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The work $W_{\text{by }\vec{\bf F}}$ by some force $\vec {\bf F}$ over a path $C:\vec {\bf r}(t)$ is $$W_{\text{by }\vec{\bf F}}=\int_C \vec {\bf F} \cdot \mathrm d\vec {\bf r}.$$

Recall $\vec {\bf{v}} = \dfrac{\mathrm d \bf\vec {r}}{\mathrm dt}$, then, for some time interval from $t_0$ to $t$, the work is

$$W_{\text{by }\vec{\bf F}}=\int_{t_0}^{t} \vec {\bf F} \cdot \vec {\bf v}\ \mathrm dt'$$

where the $t'$ is used simply for notation (since we can't have the variable of integration as an integral bound).

We know power is the time derivative of work, therefore,

$$P=\dfrac{\mathrm d}{\mathrm dt}\int_{t_0}^{t} \vec {\bf F} \cdot \vec {\bf v}\ \mathrm dt',$$

and, by the fundamental theorem of calculus, we have that $$P=\vec {\bf F} \cdot \vec {\bf v}.$$

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  • $\begingroup$ So I understand that this formula apply even when the force is not constant, am I right?, thank you for your help. $\endgroup$ – Sohaib Ali Alburihy Apr 29 at 0:33
  • $\begingroup$ @SohaibAliAlburihy yes. we never had to impose any restrictions on the force $\endgroup$ – user256872 Apr 29 at 1:22

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