Help on force, work, velocity and friction?

AI Thread Summary
A car moving at a constant speed of 5.0 m/s along a 6.0 km road faces a frictional force of 150 N/m. To calculate the work done to maintain this speed, the frictional force must be multiplied by the distance, leading to a total work of 900,000 J or 900 kJ. However, confusion arises regarding the units of frictional force and the discrepancy with the answer sheet, which states 900 J. The discussion emphasizes that the frictional force is defined per unit length, and while power is typically associated with velocity, the problem specifically asks for work done. The participants seek clarification on the correct interpretation of the problem and the calculations involved.
Tuvshee
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Homework Statement


It says: A car moves at a constant speed of 5.0ms-1 along a 6.0km road against frictional force of 150N per meter. How much work has to be done by the car to maintain its constant speed?


Homework Equations


W = F*d


The Attempt at a Solution


So I just did 150N * 6000m = 900000J = 900kJ

But what about the 5ms-1? Is that important? I'm guessing it is since it is mentioned, but I don't know how to apply it. And the answer sheet says 9000kJ (and not 900kJ). Please help me? I'm confused, thank you.
 
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Tuvshee said:

Homework Statement


It says: A car moves at a constant speed of 5.0ms-1 along a 6.0km road against frictional force of 150N per meter. How much work has to be done by the car to maintain its constant speed?


Homework Equations


W = F*d


The Attempt at a Solution


So I just did 150N * 6000m = 900000J = 900kJ

But what about the 5ms-1? Is that important? I'm guessing it is since it is mentioned, but I don't know how to apply it. And the answer sheet says 9000kJ (and not 900kJ). Please help me? I'm confused, thank you.

The frictional force is 150 N/m so if you multiply that by the 6000 m you will get the frictional force experienced throughout the entire 6000 m. In order to maintain the 5 m/s you'd need to get rate of work that needs to be done to overcome the frictional force.
 
rock.freak667 said:
The frictional force is 150 N/m so if you multiply that by the 6000 m you will get the frictional force experienced throughout the entire 6000 m. In order to maintain the 5 m/s you'd need to get rate of work that needs to be done to overcome the frictional force.

Ok I get that. But if the unit of frictional force is N/m, then it's not a force anymore, is it? Because unit for force is N and not N/m. Anyways: force would be:
150N/m * 6000m = 900000N.

v = 5m/s
N = 900000N

Then? Which formula should I use?
 
Have you tried to approach the problem using power?
Whenever I see velocity mentioned, I think of power.
 
Tuvshee said:
Ok I get that. But if the unit of frictional force is N/m, then it's not a force anymore, is it? Because unit for force is N and not N/m. Anyways: force would be:
150N/m * 6000m = 900000N.

v = 5m/s
N = 900000N

Then? Which formula should I use?

They didn't give you the force for the entire length, they just gave you the force per unit length, so it is still a force just defined per unit length.


Normally, I'd think they were asking for power rather than the work done.
 
rock.freak667 said:
They didn't give you the force for the entire length, they just gave you the force per unit length, so it is still a force just defined per unit length.


Normally, I'd think they were asking for power rather than the work done.

Oh ok :) well they were asking for the work done...oh and correction. The answer sheet says 900J. But it's still different than my result (900kJ). And I'm still stuck with no clue. :( And I don't think power is involved, because we haven't covered the topic yet.
 
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