Let's assume that a car is travelling at a constant velocity (V) on a horizontal flat plane , hence the force of friction is balanced by the force applied by the engine.
The power which the engine exerts upon the car is given by the equation $Fengine×V $.
Now let's assume that the same car now moves up an incline plane at a constant velocity, with the engine exerting the same amount of power.
However since the force exerted by the engine will now increase , the constant velocity at which the car travels up the slope should decrease if the power is to remain constant.
Hence I would like an understanding , preferably newtonian as to what caused such a decrease in speed? Aren't the forces balanced , does this not go against newtons first law , since in this case no unbalanced force acted upon the object to cause a change in motion. My intuition tells me that the time which it took the engine to exert a force to balance $Fparrallel$ and $Ffrict$ was the time in which the reduction in velocity took place , but I would like some confirmation .
[Physics] what causes a decrease in velocity
newtonian-mechanicspowerwork
Best Answer
You're right that the reduction in velocity takes some time. Let me spell out in more words/equations what you seem to already understand!
You're demanding that the power, $P=F_{\text{engine}}v$ is constant. You also have some $F_{\text{friction}}$, the force due to rolling friction, and $F_{\text{gravity}}$, the force pulling the car backwards along its axis of motion due to gravity.
From free body diagrams, $F_{\text{friction}}=\mu g M \cos(\theta)$, where $\theta$ is the angle of the slope, $\mu$ is the coefficient of rolling friction, and $gM$ is the normal force due to gravity. The force of gravity along the direction of motion is $F_{\text{gravity}}=g M \sin(\theta)$.
If the car's velocity is in equilibrium, then we have:
\begin{align*} 0&=F_{\text{engine}}-F_{\text{friction}}-F_{\text{gravity}}\\ &=\frac{P}{v}-\mu g M\cos(\theta)-g M\sin(\theta) \end{align*}
So
$$v=\frac{P}{g M(\mu\cos(\theta)+\sin(\theta))}$$
This $v$ is the equilibrium velocity. If $\theta=0$ this just gives the velocity needed to push a block at constant power. If $\theta=\pi/2$ this just gives the velocity needed to lift a block at constant power. For a low coefficient of rolling friction, pushing is easier than lifting, and the higher theta is the lower the velocity for constant power is.
I think you've grasped all this intuitively in fewer words and equations! Obviously if someone is going straight and starts going up a hill without stepping more on the gas pedal, they're going to slow down over time.
There are a lot of assumptions made, however. For example, the relation between $F_{\text{engine}}$ and the power $P$ is probably wrong. You would expect the force that an engine delivers to depend on the velocity, the load placed on it, and all sorts of other things. My point is: this is kind of a weird constraint to put on an engine (for example, it breaks down when the car starts from a standstill and $v=0$). If you're working for BMW then you need a better model, but it suffices to us. We have a description showing that the car has to slow down.
Unfortunately it's harder to figure out how the slow-down happens. There is in fact an unbalanced force, so we have to start plugging stuff into Newton's law $F=ma$. We have to worry about the equation
$$ma=\frac{P}{v}-\mu g M\cos(\theta)-g M\sin(\theta)$$
which is an ugly, ugly thing. It's so ugly because it's nonlinear. The $v$ on the bottom completely screws up any non-calculus based methods you've learned so far (You wind up with an ordinary differential equation of the form $y''(t)=A/y'(t)+B$, where $y''$ is acceleration and $y'$ is velocity). It's easier to resort to numerics to solve it.
Let's look at the following situation: We have a 1 metric ton car which is being pulled by 1000 watts of power (this is a bit over 1 horsepower). It goes on flat ground at constant velocity for a bit, then goes up a three degree slope. That is: $\mu=0.01$, $g=9.81 \text{m}/\text{s}^2$, $M=1000\text{kg}$, $P=1000\text{watts}$, $\theta=\pi/60$. The graph of velocity over time looks as follows:
It starts at about ten meters per second and travels along at constant velocity. When it hits the slope, it takes about twenty seconds to slow all the way down to a bit less than two meters per second.
The graph was generated with the following Mathematica code. (Note: HeavisideTheta[t] returns 0 if t<0 and 1 if t>0)