How does doubling the force affect the motion of a box being pushed?

[jesuslovesu]

Homework Statement

You exert a constant horizontal force on a large box. As a result the box moves across a horizontal floor at a constant speed v0.

If you double the constant horizontal force on the box, how would the box then move?

Homework Equations

F = ma = m dv/dt

The Attempt at a Solution

Well, this question is a tad confusing to me, if the box is moving at a constant speed v0 then I have to assume that I would just push the box for an instant & then get it to move at a constant v0.

If I were to double the force, then I would think that the speed would then be 2v0, however this is not the case... (dv/dt is proportional to F)

Can anyone explain how to determine the motion?

 

[Doc Al]

Well, this question is a tad confusing to me, if the box is moving at a constant speed v0 then I have to assume that I would just push the box for an instant & then get it to move at a constant v0.

You are misinterpreting the problem. The only reason the box continues moving at constant speed is because you keep pushing with the constant horizontal force. Hint: What's the net force acting on the box? What other force must be acting on the box?

 

[jesuslovesu]

Thanks for your reply
Alright so if I draw a free body diagram and add in friction I can see
ma = F - Ff
Since the velocity is constant a = 0 so
F = Ff

So if I double F to 2F, then it's accelerating?
2F - Ff = ma
F = ma
a = F/m ?

 

[Doc Al]

Very good!