If a bullet penetrates a bag, how come the repulsive force is constant? I was doing a question on energy and forces, and it goes as follows:(Doesn't require knowledge of calculus):

If a bullet with velocity $v$ penetrates a bag upto a distance $x$,
  then find the distance which bullet penetrates if velocity becomes
  $2v$.

I tried to solve it, like this: 
$
1/2mv^2 = -F * x$ because energy is conserved and retarding force given by bag would be negative.
So I multiplied both sides by 4, because velocity becomes 2 times, and the square makes it 4. Then, on the R.H.S of the equation, I got the distance to become 4 times, if force remains constant in both cases. And, my answer came out to be correct. However, I was thinking why would force be constant. By Newton's third law of motion, bullet with higher velocity will provide a greater force, but we are taking force to be constant. And, what are the pairs of equal and opposite forces in this case? If second bullet provides a greater force, shouldn't the bag provide the same greater force?
 A: The force is most likely not constant but your teacher wanted you to be able to do the problem in not too long a period of time.
A: I know where you got your working (!) equation, but you should start with a good concept:
$$\Delta KE = \int_{x_1}^{x_2}\vec{F}\mathrm{d}x.$$
And the force is opposite the propagation direction: $$0-1/2 mv^2 = \int_0^X|F|(-\hat{i})\mathrm{d}x$$
That takes care of your negative sign problem. Never simply ignore negative signs!

However, I was thinking why would force be constant.

Good question! It shouldn't be for most stuffing that would be in the bag. @Farcher gave a good answer for this. Actually, the retarding force would depend on the speed of the bullet either $F\propto v$ or $F\propto v^2$ or a linear combination of the two. Then you would have to integrate $v$ or $v^2$ along $x$. Assuming a constant force makes the integral easy.

By Newton's third
  law of motion, bullet with higher velocity will provide a greater
  force, but we are taking force to be constant.

What does Newton's 3rd Law have to do with it? The higher velocity simply means more encounters per second with particles in the bag. We interpret that as a higher force because of the greater acceleration magnitude.

And, what are the pairs of equal and opposite forces in this case?

If the bullet and a stuffing particle touch, they exert forces on each other. But the only force you need is the force on the bullet. 

If second bullet provides a greater force, shouldn't the bag provide the same greater force?

Yes. That means that a constant force assumption doesn't tell the correct story. And the work integral isn't the way to solve this. The impulse integral works better: $$\int \mathrm{d}\vec{p}=\int\vec{F}\mathrm{d}t$$
A: The bag can only give so much force to push against the bullet - it is limited. A consequence of this is that parts of the bag will be accelerated and move - the bag may 'explode'.
This is a bit like the problem of the fly and the train. If a train hits a fly the fly does not stop the train because it cannot provide enough reaction to do that. 
As the fly is hit by the train the force of the train accelerates the fly until they are travelling at the same speed.
I imagine your bag has some stuffing that gets accelerated like the fly.
