We know that force is inversely proportional to the velocity when power is constant. Can someone provide me with a real-life example where an increase in the net force acting on a body actually reduces its velocity so as to keep the power constant? (I think I'm having some conceptual dilemma here).
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2 Answers
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The examples I can think of revolve around the idea that for the same amount of power, you can get more Force but you pay the penalty of a slower speed.
Gears and power transmission: for a Car, at a lower gear you get more torque(force) but go slower. Similar things for Bikes at lower vs higher gears. Switching gears can be done real time and your torque vs angular velocity will change. Consider the case of the bicycle, When the gear ratio goes down and the F (Torque) goes up, the v (𝜔) actually goes down. In other words as it gets harder to pedal the pedalling speed goes down NOT the speed of the bicycle. The speed you can turn the crank goes down as the force goes up. The $v$ Is the velocity that you can apply the Force that changes not necessarily the velocity of the system as a whole.
Pulley system : you can use a pulley system to greatly magnify your force with the same power output but whatever you are pulling/lifting moves slower. If you remove one of the wheels in the pulley system , it will take more force but the movement will go faster.
Note that all the above examples are also closely related to multiplying the force you apply via leverage, but revolve around the idea that Energy (or for Power, Energy/time) is conserved.
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$\begingroup$ Your examples are correct, but maybe I didn't do a good job explaining my dilemma. When we say P = Fv, if F means the frictional force opposing the motion, then F and v do seem inversely proportional. But if F means the force that pushes the body in the direction of motion, then v seems to increase with F. So what exactly are F and P here? Does 'P' mean the change in KE/time of the body or the change in energy/time transferred to the body to oppose or aid its motion? $\endgroup$ Commented Jul 16, 2021 at 11:24
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$\begingroup$ I have edited my answer based on you comment: I think I understand your question a bit better. Consider the case of the bicycle (please forgive the usage of Torque and $\omega$ Instead of F and v), When the gear ratio goes down and the F (Torque) goes up, the v ($\omega$) actually goes down. In other words as it gets harder to pedal the pedalling speed goes down NOT the speed of the bicycle. The speed you can turn the crank goes down as the force goes up. It’s the speed that you can apply the Force that changes not the velocity of the system. Does that make it clear? $\endgroup$ Commented Jul 16, 2021 at 15:26
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1$\begingroup$ I didn’t really answer your follow up question. F is the force you apply on the pedals of the bike , v is the speed that you can press the pedals of the bike. P is the power (KE/time) that you impart on the bike. If your power is constant then pedalling speed is inversely related to Force you need to apply $\endgroup$ Commented Jul 16, 2021 at 15:34
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$\begingroup$ Thanks. I think it is clear now. I got confused because I took 'v' as the velocity of the bike instead of the pedaling speed. $\endgroup$ Commented Jul 17, 2021 at 10:43
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You might have a misconception here, instead of this:
"increase in the net force acting on a body actually reduces its velocity so as to keep the power constant?"
It's best to think of it like this:
An increase in velocity reduces the force available
or
A reduction in velocity can increase the force available.
It comes from work done = Force x distance and dividing by time on both sides
$$W=Fd$$
$$\frac{W}{t}=F\frac{d}{t}$$
$$P=Fv$$
Let's imagine someone trying to drag something along the ground - and that they can only have a certain power output. If they tried to drag it fast, they would have less force available, so could drag a smaller object fast (so $Fv$ is constant), but if they dragged a heavier object, maybe they could do so, but more slowly.
answered Jul 16, 2021 at 8:24
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$\begingroup$ This is actually where I'm having trouble with. When we say P = Fv, if F means the frictional force opposing the motion, then F and v do seem inversely proportional. But if F means the force that pushes the body in the direction of motion, then F and v seem directly proportional. So what exactly is F and P here? Does 'P' mean the change in KE/time of the body or the change in energy/time transferred to the body to oppose or aid its motion? $\endgroup$ Commented Jul 16, 2021 at 11:22
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$\begingroup$ *v seems to increase with F. (not directly proportional) $\endgroup$ Commented Jul 16, 2021 at 11:29
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$\begingroup$ $F $is the force provided by the person (or car) , $v$ is the velocity of the person (or car) $Fv$ is constant ($=P$), so $F$ is inversely proportional to $v$ $\endgroup$ Commented Jul 16, 2021 at 12:13
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$\begingroup$ If 'F' is the force applied by the person to drag a block along the ground, then shouldn't the velocity of the block increase with the increase in F? $\endgroup$ Commented Jul 16, 2021 at 12:43
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$\begingroup$ Let's go back to the car then, in case the block example is confusing. It's about the limitations of the car engine, that can only provide a certain power output. Let's say it's limited to 100W. At 10m/s it can provide 10N of forward force. (it goes 10m in 1 second and also from W=Fd, you can see it can provide 100J in the second). If it went at 20m/s it could only provide 5N (it goes 20m in 1 second, from W=Fd it provides 100J again in the second). So for a constant (limited) power output, best to think that the force that the engine can provide is inversely proportional to velocity. $\endgroup$ Commented Jul 16, 2021 at 12:58