What does the equation $P=\vec{F}\cdot \vec{v}$ really mean? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-20T05:02:58Z https://physics.stackexchange.com/feeds/question/339100 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/339100 1 What does the equation $P=\vec{F}\cdot \vec{v}$ really mean? LM26 https://physics.stackexchange.com/users/152979 2017-06-13T09:52:28Z 2017-06-14T08:38:33Z <p>Let's say that we have a car travelling at a constant velocity (V) on a flat plane , let's assume that it's applied force or the force applied from its engine ( Fapp ) balances the frictional force ( Ffrict ) , hence the power due to Fapp is equal to Fapp×V , now let's assume that the same car now moves up an incline with the same engine power as previously stated and at a constant velocity ...but this time it will travel at a much lower constant velocity since the force which the engine exerts is much higher now in order to balance Fparalell and Ffrict , but what I don't understand is what caused this reduction in velocity , if the engine applies an equal force to that of the counteracting forces shouldn't that mean that there is no net force hence no change in velocity ,also if p=FV and if we now apply a greater force for the same velocity V why does the power increase , I know the answer can be simply explained using mathematics but my intuition says that power is the rate at which kinetic energy is transferred to a particle , so why would a greater force at the same constant velocity mean that you are transferring a greater amount of KE per unit time ? </p> https://physics.stackexchange.com/questions/339100/-/339103#339103 0 Answer by Yashas for What does the equation $P=\vec{F}\cdot \vec{v}$ really mean? Yashas https://physics.stackexchange.com/users/60891 2017-06-13T09:55:54Z 2017-06-13T09:55:54Z <p>The formula can be understood in two ways: analytically and mathematically,</p> <p>The mathematical derivation is as follows:</p> <p>$$dW = F.dx$$</p> <p>$$P = \frac{dW}{dt}$$</p> <p>If the force remains constant and acts in the direction of the displacement, then:</p> <p>$$P = F\frac{dx}{dt} = Fv$$</p> <p>Power by definition is the rate of doing work. Work is done when there is a displacement. Therefore, the object must have a velocity to do work.</p> https://physics.stackexchange.com/questions/339100/-/339110#339110 2 Answer by Farcher for What does the equation $P=\vec{F}\cdot \vec{v}$ really mean? Farcher https://physics.stackexchange.com/users/104696 2017-06-13T10:16:31Z 2017-06-14T08:38:33Z <blockquote> <p>but my intuition says that power is the rate at which kinetic energy is transferred to a particle , so why would a greater force at the same constant velocity mean that you are transferring a greater amount of KE per unit time?</p> </blockquote> <p>Your definition of power in terms of rate of transfer of kinetic energy cannot be correct because the kinetic energy is constant if the velocity is constant. </p> <p>Going up the slope there are forces $F_{\rm parallel}$ and $F_{\rm friction}$ acting on the car.<br> Since the car is moving at constant velocity there must be another external force $F_{\rm external}$ acting on the car which is equal in magnitude and opposite in direction to the sum of those two forces.<br> This means that the net force on the car is zero and the net work done on the car is zero which means that its kinetic energy stays constant.</p> <p>That "other" external force on the car does work to move the car up the slope and the power delivered is $F_{\rm external} \,v$ where $v$ is the speed (velocity) of the car. </p> <p>The powers related to the other two forces $F_{\rm parallel}\, v$ and $F_{\rm friction} \, v$ end up as an increase in the gravitational potential energy of the car-Earth system and heat. </p> <p>On the flat the $F_{\rm parallel}$ force is zero and so the "other" external force is smaller and has to deliver less power to keep the car moving at the constant velocity. </p> <hr> <p><strong>Update</strong> as a result of a question from the OP.</p> <p>If the velocity is constant then the acceleration is zero.<br> Let the velocity up the slope be $v_{\rm up}$ and velocity on the flat be $v_{\rm flat}$.<br> If the power is constant the equation the power going up the slope to the equal power on the flat gives:</p> <p>$(F_{\rm parallel} + F_{\rm friction})\, v_{\rm up} = F_{\rm friction}\, v_{\rm flat}$ </p> <p>which means that $v_{\rm flat} &gt; v_{\rm up}$</p> https://physics.stackexchange.com/questions/339100/-/339132#339132 2 Answer by Luke Pritchett for What does the equation $P=\vec{F}\cdot \vec{v}$ really mean? Luke Pritchett https://physics.stackexchange.com/users/89597 2017-06-13T12:47:29Z 2017-06-13T12:47:29Z <p>When you see an equation that involves force it's important to always ask "which force?" Here, the meaning of $P$ depends on which $F$ you are talking about.</p> <p>If you have an object with velocity $\vec{v}$ then the rate of its kinetic energy its $\frac{dK}{dt} = \frac{d}{dt}(\frac{1}{2}m\vec{v}\cdot\vec{v}) = m\vec{a}\cdot \vec{v} = \vec{F}_{\text{net}}\cdot \vec{v} = P_{\text{kinetic}}$. Note that if there is a non-zero $P_{\text{kinetic}}$ then $F_{\text{net}}$ is automatically non-zero, and so the velocity is changing.</p> <p>But this is kind of a limited view because it doesn't let us break apart contributions from different forces, and it doesn't allow us to consider forces that feel like they are doing work even when the velocity doesn't change (eg an engine driving a car at a constant speed along a flat road).</p> <p>So let us consider an object with both external and conservative forces acting on it. Conservative forces are ones that have an associated potential energy, with the relation $\vec{F}=-\nabla V$. Then we can define the total energy $E = K + V$ and find its rate of change: $$P_\text{total} = \frac{dK}{dt}+\frac{dV}{dt} = \vec{F}_\text{net}\cdot \vec{v}+\vec{v}\cdot \nabla V = (\vec{F}_\text{net} - \vec{F}_\text{conservative})\cdot \vec{v} = \vec{F}_\text{external}\cdot \vec{v}$$</p> <p>This helps us to break apart forces a bit better. We could consider the contribution of each individual force to the total rate of change of energy. Though note that this is not technically a different equation than the one above: we've just put some of the terms on a different side.</p> <p>For example, consider a mass being pushed up an inclined plane at a constant speed $v$. Its potential energy is changing at a rate of $mgv\sin\theta$, and so that is the rate of energy that must be expended by external forces.</p>