How hard must you push a box to move it with a constant speed?

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Homework Help Overview

The discussion revolves around determining the force required to push a 2kg box with a coefficient of kinetic friction of 0.2 so that it moves at a constant speed of 5m/s. Participants are exploring the implications of constant speed and the role of friction in this context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the force of kinetic friction and its implications for maintaining constant speed. Questions arise about the initial conditions necessary to reach the speed of 5m/s and the effects of applying a force greater than the frictional force.

Discussion Status

There is an ongoing exploration of the problem, with some participants agreeing on the calculated force of 4N while questioning the clarity of the problem statement. Others are considering the implications of applying forces greater than 4N and how that relates to acceleration.

Contextual Notes

Participants note the lack of information regarding the initial speed of the box and the time required to reach the constant speed of 5m/s, which complicates the analysis.

daiea08
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Homework Statement


How hard do you need to push a 2kg box which has .2 coefficient of kinetic friction so that it moves with a constant speed of 5m/s?


[

The Attempt at a Solution


(I'm calling g 10m/s2 not 9.8)
F(kinetic friction)=(.2)Fn=(.2)mg=(.2)(20N)=4N
so...since the acceleration is zero, the force must equal 4N..so 4N is the answer? If I'm correct then in the "5m/s" just a deterrent? Is the key point "constant speed?"
 
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daiea08 said:

Homework Statement


How hard do you need to push a 2kg box which has .2 coefficient of kinetic friction so that it moves with a constant speed of 5m/s?


[

The Attempt at a Solution


(I'm calling g 10m/s2 not 9.8)
F(kinetic friction)=(.2)Fn=(.2)mg=(.2)(20N)=4N
so...since the acceleration is zero, the force must equal 4N..so 4N is the answer? If I'm correct then in the "5m/s" just a deterrent? Is the key point "constant speed?"

The final acceleration is zero, but how exactly does the block speed up to get to 5m/s?How could you work out the force it needs to accelerate?
 
daiea08 said:
so...since the acceleration is zero, the force must equal 4N..so 4N is the answer? If I'm correct then in the "5m/s" just a deterrent? Is the key point "constant speed?"
I'd say you were correct. The 5 m/s was just to throw you off!

The problem could be worded better as it implies you are supposed to figure out how to get it to 5 m/s from some unspecified prior condition. But you aren't given any info as to its initial speed or the time needed to get the final speed, so you can't answer that.
 
Doc Al said:
I'd say you were correct. The 5 m/s was just to throw you off!

The problem could be worded better as it implies you are supposed to figure out how to get it to 5 m/s from some unspecified prior condition. But you aren't given any info as to its initial speed or the time needed to get the final speed, so you can't answer that.

Yeah, that was my initial though too, and tbh I'm a little rusty with this stuff, but what would happen if a force in excess of 4N was applied wouldn't its velocity increase i.e accelerate?
 
malty said:
Yeah, that was my initial though too, and tbh I'm a little rusty with this stuff, but what would happen if a force in excess of 4N was applied wouldn't its velocity increase i.e accelerate?
Sure.
 

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