Understanding the Relationship Between Force and Motion in Physics

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Homework Help Overview

The discussion revolves around the relationship between force, power, and motion in physics, specifically focusing on the equations relating power to work done and the forces acting on a moving car. Participants explore the implications of different forces, including driving and frictional forces, in the context of a homework problem involving a car's power and speed.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of power in relation to force and speed, questioning whether the force in the equation refers to the driving force or the opposing frictional force. There are inquiries about the role of kinetic energy in calculating frictional force and the implications of constant speed on net force.

Discussion Status

The conversation is active, with participants providing insights into the nature of forces and power. Some guidance has been offered regarding the relationship between applied and dissipated power, and there is an ongoing exploration of how to approach the homework problem without reaching a consensus.

Contextual Notes

Participants note the constraints of the homework problem, including the given power, speed, and mass of the car, while questioning the relevance of acceleration and the types of forces considered, such as friction and air drag.

jeekeshen
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Am getting really confuse with some force related equation..


Power = Workdone / Time

= Force X Speed

BUT THIS FORCE is it the one that makes the object move or the friction force( the one opposing the motion )

Because one of my homework makes me think that it's the frictional force, pleasezz help
 
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Power = Force*Speed is correct. However, we can see that energy of a constant force applied over a distance d is F*d. If this force is applied over the distance d for a time t, then the average speed will be d/t. Thus, this you could say that P = F*v. More importantly, if we think of the force being applied over an infinitesimal distance dx, for an infinitesimal amount of time, dt, then we will find that P(t)=F(t)*(dx/dt)=F(t)*v(t). So this definition is true at any given time t.
 
Welcome to PF!

Hi jeekeshen! Welcome to PF! :smile:
jeekeshen said:
BUT THIS FORCE is it the one that makes the object move or the friction force( the one opposing the motion )

Because one of my homework makes me think that it's the frictional force, pleasezz help

We usually talk about the power of a machine, so yes, that means using the force that makes the object move. :smile:

However, if the object isn't accelerating, that force will be the same as the friction force (well, ok, the opposite! :rolleyes:), and often the friction force is easier to calculate :wink:

I assume your homework question tells you only the speed the mass and the coefficient of kinetic friction ?
 
By definition, *any* force can do work. If the power is calculated for the friction force, it just gives the value of the quantity of energy dissipated by friction per second. We calculate the frictional power for determining the brake power of the engine-the one obtained at the flywheel.
 
The real Question : The power of the car is 36.6 kw and speed is 32m/s
the mass of the car is 720 kg...

Calculate the magnitude of the external force opposing the motion of the car...??
 
Well, in general you can have a power for each force - or more precisely, each force either applies or dissipates some power. So if you have a force that is pushing the car, like the force from the engine, you can calculate the power applied by that force, and if you have a force that is resisting the push, like friction, you can calculate the power dissipated by that force. An "applied" power will be positive, and a "dissipated" power will be negative; and if you add up the (positive) applied power from the engine and the (negative) dissipated power from the friction, you will get the net power exerted to move the car. That will be the same result you would get if you calculated the power exerted by the net force (engine minus friction) acting on the car.

In the question you posted, it looks like they give you the power exerted by the engine, and they want you to figure out the frictional force.
 
Yes but i have its kinetic energy also... would this help to calculate the frictional force ?
 
jeekeshen said:
… speed is 32m/s
the mass of the car is 720 kg...
jeekeshen said:
Yes but i have its kinetic energy also... would this help to calculate the frictional force ?

If the speed is constant, why would KE come into it? :confused:
 
jeekeshen said:
The real Question : The power of the car is 36.6 kw and speed is 32m/s
the mass of the car is 720 kg...

Calculate the magnitude of the external force opposing the motion of the car...??
Since you're given the power and speed of the car, use that to figure out the effective force providing that power. Then ask: What's the net force on the car? (Is it accelerating?)
 
  • #10
No not accelerating... the net force will perhaps be; using F= Ma where f is the resultant force, 720 X 9.8 = net force...
 
  • #11
jeekeshen said:
No not accelerating... the net force will perhaps be; using F= Ma where f is the resultant force, 720 X 9.8 = net force...
If it's not accelerating, then the acceleration = 0. So what's the net force?
 
  • #12
jeekeshen said:
720 X 9.8 = net force...

The car isn't in free fall!
 
  • #13
O yea acceleration is therefore zero WTF
 
  • #14
jeekeshen said:
O yea acceleration is therefore zero
Right, the acceleration is zero and so is the net force. So the two forces acting on the car--the driving force and the opposing force--must be equal and opposite. The power equation will allow you to figure out the driving force.
 
  • #15
JUST for general knowledge,
is this question valid for 4 driving wheels( i mean, some of the cards(most) have two driving wheels(i hope that's who you call them) , in the front of the car where there the engine is )' but will it be valid?
or the question is for any force, not just friction with the ground, could be air drag as well?right?
thx
 

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