Why Does Doubling the Force More Than Double the Acceleration?

[Quark Itself]

Normally we are used to the fact that if "a" in F=ma doubles, then Fnet as well should double and if "m" is halved then F is also doubled.
Can anyone explain why it isn't so in this case?

"A horizontal force accelerates a box from rest across a horizontal surface (friction is present) at a constant rate. The experiment is repeated, and all conditions remain the same with the exception that the horizontal force is doubled. What happens to the box's acceleration?"

Apparently, the answer is that Acceleration is increased to more than twice its original value.

 

[nonequilibrium]

If we double Fnet, a will indeed double (taking mass as constant). However, here we are doubling a force, but not the net force. What force other than us pushing is there? How does that behave when we push onto the box?

 

[Dadface]

F refers to the resultant force,the horizontal force referred to in the question minus the frictional force.In the question the horizontal force doubles but friction remains constant so the resultant force more than doubles.

 

[physicsworks]

Because doubling pushing force in this case doesn't mean doubling total force which is pushing force plus frictional force (vectors).

 

[arildno]

QUARK:
We can calculate this as follows:

Let H be the originally applied force, F the frictional force, m the mass, "a" the original acceleration and A the acceleration in the second experiment.
(All numbers positive)

For the two cases then, we have:

H-F=m*a (1)

2H-F=m*A (2)

(2) can be eqivalently written as:

2H-2F+F=m*A, that is, using (1):

2m*a+F=m*A

That is:
A-2*a=F/m>0, i.e, A>2*a

 

[Quark Itself]

If we double Fnet, a will indeed double (taking mass as constant). However, here we are doubling a force, but not the net force. What force other than us pushing is there? How does that behave when we push onto the box?

So in other words, Fapp is doubled but Fnet in comparisson between the experiments is more than doubled? If I've understood it correctly.
But how would you put it in terms of Algebraic equations?

I have tried these:
Fapp - μmg = ma1
2Fapp - μmg = ma2

but, that's as far as I get.

 

[Quark Itself]

QUARK:
We can calculate this as follows:

Let H be the originally applied force, F the frictional force, m the mass, "a" the original acceleration and A the acceleration in the second experiment.
(All numbers positive)

For the two cases then, we have:

H-F=m*a (1)

2H-F=m*A (2)

(2) can be eqivalently written as:

2H-2F+F=m*A, that is, using (1):

2m*a+F=m*A

That is:
A-2*a=F/m>0, i.e, A>2*a

Thank you for your quick reply !
Thanks to everyone to be precise !

 

[arildno]

Thank you for your quick reply !

You're welcome!

 

[zzoo4]

This might be a random question.. but where did you get this question?

 

[Ken G]

I have tried these:
Fapp - μmg = ma1
2Fapp - μmg = ma2

Note that it is sometimes a useful device to take a problem and push it to some kind of extreme. Here, the extremes are to make the horizontal force so huge that friction is negligible, or to make it so small that it is barely larger than friction (kinetic friction is often assumed to be a constant force, so let's do that, even though I doubt it's a very good approximation). If the horizontal force is huge, doubling it will double the acceleration, since friction is treated as negligible. But if the horizontal force is so barely larger than friction that the mass is barely accelerating at all, and we double the horizontal force, it is obvious that the tiny acceleration will much more than double, since it can be made arbitrarily small at first (with a force arbitrarily larger than friction) yet still have a good final acceleration (indeed, essentially the same final acceleration).