# Why Does Doubling the Force More Than Double the Acceleration?

• Quark Itself
In summary, in the given scenario, if the horizontal force is doubled, the acceleration of the box will increase to more than twice its original value. This is because doubling the pushing force does not necessarily mean doubling the net force, as there may be other forces (such as friction) acting on the box. This can be mathematically represented by the equation A-2*a=F/m>0, where A represents the acceleration in the second experiment, a represents the original acceleration, and F/m represents the net force.
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.

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?

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.

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

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

mr. vodka said:
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.

arildno said:
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 !

Quark Itself said:
Thank you for your quick reply !
You're welcome!

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

Quark Itself said:
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).

## What is the meaning of F=ma?

F=ma is a fundamental equation in physics known as Newton's Second Law of Motion. It states that the force (F) applied to an object is equal to its mass (m) multiplied by its acceleration (a).

## Why does F=ma not make sense?

F=ma is a simplified equation that only applies to objects moving at a constant acceleration. It does not account for other factors such as friction, air resistance, and the shape of an object. In real-life situations, F=ma may not accurately predict the motion of an object.

## What happens if the mass in F=ma is zero?

If the mass (m) in F=ma is zero, then the equation becomes F=0. This means that no matter how much force is applied, the object will not accelerate. This is because the mass of an object determines its resistance to acceleration.

## What is the unit of measurement for F=ma?

The unit of force in F=ma is Newtons (N), the unit of mass is kilograms (kg), and the unit of acceleration is meters per second squared (m/s^2). Therefore, the unit for F=ma is kg*m/s^2.

## Can F=ma be applied to all types of motion?

No, F=ma only applies to objects moving at a constant acceleration. For other types of motion, more complex equations, such as those in calculus, are needed to accurately describe the motion of an object.

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