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

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To move a 2kg box with a coefficient of kinetic friction of 0.2 at a constant speed of 5 m/s, a force of 4N is required to overcome friction. The problem's mention of 5 m/s is considered a distraction, as the key focus is on maintaining constant speed, which implies zero acceleration. Without information on the box's initial speed or the time to reach 5 m/s, the problem lacks sufficient detail for a complete analysis of acceleration. If a force greater than 4N is applied, the box would indeed accelerate. Understanding the distinction between constant speed and acceleration is crucial in solving such physics problems.
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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