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If an object is moving at a constant speed the force of friction must equal the applied (horizontal) force, and for it to be accelerating or decelerating, the force of friction and the applied force must be unequal. Also, I know that $f = \mu N$.

This is what I do not understand: if the applied force is greater than the friction, wouldn't that mean the object would continue to accelerate infinitely? Shouldn't the friction force change to equal the applied force to prevent this? If so, how do I work out how the friction force $f$ changes?

Here is a sample situation: say I have a box with mass $10$ kg and I apply a horizontal force $50$ N, and the coefficient of kinetic friction is $0.5$. How long does it take for the box to finish accelerating and reach a constant velocity? Then, if I increase the force to $60$ N, how long does it take to reach constant velocity again?

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This might be more detailed than you want; I apologize in advance.

There are two forms of friction:

  1. static friction The force of friction exerted on an object when it is at rest.

  2. kinetic friction The force of friction exerted on an object when it is in motion.

These two forms of friction have qualitatively properties. Specifically, the force of kinetic friction depends only on the magnitude of the normal force $F_N$ exerted on the moving object and the coefficient of kinetic friction $\mu_k$ of the surface on which it is moving. In fact, as you point at the magnitude of the force of kinetic friction as given by $$ F_k = \mu_k F_N $$ The force of static friction, on the other hand, changes depending on the other external forces on the object.

To understand why, think of a box sitting still on a horizontal table. The box will not feel a friction force in the absence of any other force (if it did, then it would accelerate). However, if you start exerting a small enough force on the box, it still will not move, and in this case, the static friction force is exactly counterbalancing the force you exert. If you push hard enough, however, the box will eventually start sliding. This illustrates that the static friction force can have any value between zero, and some maximum which turns out to be given by $\mu_s F_N$ where $\mu_s$ is the coefficient of static friction. Mathematically, this can be expressed by the following equation: $$ F_s \leq \mu_s F_N $$ where $F_s$ is the magnitude of the static friction force.

Having said all of this, let me reiterate that kinetic friction always has the magnitude $\mu_k F_N$ regardless of the state of motion of the object. If you continuously push an object with a force greater than this value, then it will keep accelerating forever. In order for the acceleration to halt, you would need to reduce the applied force so that it equals the force of static friction.

Lastly, given an applied force $F$, the acceleration of the object will satisfy Newton's second law which says that the net applied force equals the mass of the object multiplied by is acceleration; $$ F - F_k = ma $$ The acceleration of the object is the rate of change of its velocity, so determine the velocity as a function of time you would, in general, have to solve the following differential equation: $$ \frac{dv}{dt} = \frac{1}{m}(F - F_k) $$

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  • $\begingroup$ Thank you, this really helped clear up a few things. However something I do not understand is if you increase the force you apply to an object passed its kinetic force of friction it does not keep accelerating doesnt it just reach a higher constant veloctiy? $\endgroup$
    – arch
    Mar 10, 2013 at 21:46
  • $\begingroup$ @ρσݥzση Sure thing. If you keep exerting a force greater than the force of static friction, then no, the object will not simply reach a higher constant velocity. It's velocity will keep increasing. $\endgroup$
    – joshphysics
    Mar 10, 2013 at 21:49
  • $\begingroup$ Ok, so if I had a box and I am pulling it along by a rope, and I wanted to raise the velocity then keep it constant, would I be pulling harder to accelerate it, then softer to keep it constant? Because doing it in real life just feels like pulling it harder and then carrying on pulling it at that hardness. (Sorry for my terrible use of words, I am quite tired and can't think straight :P ) $\endgroup$
    – arch
    Mar 10, 2013 at 22:21
  • $\begingroup$ @ρσݥzση Yes exactly. $\endgroup$
    – joshphysics
    Mar 10, 2013 at 22:28
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You are actually completely right, and then at the last minute you look the wrong way!!!

Lets examine the last bit. If you were to continue applying a net force on an object for an infinite time it would indeed accelerate infinitely. There is nothing wrong here. The key point is the NET force, that is, the net amount of force, and its direction, that is applied due to the acting friction (resisting, negative) and the pushing you do(pushing, positive). Also, if you were to stop pushing on your object then your net force is only due of friction; this will cause it to slow down (accelerating, negative). LIke you initially said you gotta balance the friction and how hard you push to get that constant speed. If you don't get the balance right, and you continuing pushing forever, your speed will increase/decrease forever!!!

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This answer is probably long past its sell by date?

All you have described in your question are ideal cases with the frictional forces and the applied force constant.
If those assumptions are made then in Classical Physics the mass will continue to accelerate and not reach a terminal speed.

However when one looks at a more realistic situation with the friction force being dependent on the speed of the mass and the limitations of the device applying the force then you might understand why it is that the acceleration might not continue to be constant for all time.

The basic equation of motion is $F_{\rm applied} - F_{\rm friction} = ma$

If the left hands side remains constant and the mass $m$ remains constant then the acceleration $a$ stays constant.
Even in this situation there may be a possible constraint that have not thought of which is the power input needed to accelerate the mass.
This is $F_{\rm applied} \, v$ where $v$ is the speed of the mass.
You will note that as the speed increase so must the power input.

If $F_{\rm applied}$ is kept constant then the assumption that $F_{\rm friction}$ is constant is unlikely to hold as the speed of the mass increases.
For example the fluid around the mass, which has not been mentioned in the ideal case, might make a significant contribute to the frictional force and there are many well documented examples where the frictional force is proportional to the speed of the mass or proportional to the (speed)$^2$.
These are commonly used examples but it might be that the frictional force depends on speed in a more complex way.

So in analysing the situation when a mass reaches a terminal speed with a constant force being applied to it you must have knowledge of how the frictional forces depend on the speed of the mass.

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F=M x A= F
25 x 25=6.25

For the explanation is,the formula must be F=M x A=F

For example 25 kg. is the mass and .25.00.00 is the acceleration of the speed limit.Now you must multiply the two 25 and the answer is 6.25

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The object will continue accelerating to just below light speed (speed limit of the universe), but only in a vacuum. Unfortunately I only know a little about air resistance, but apparently the faster an object travels the more the atmosphere tries to resist the object. This is where aerodynamics comes in.

I'm sure you're familiar with the classic image of space shuttle reentry with the burning red shuttle bottom speeding through the upper atmosphere. That heat is produced by rapidly smashing into oxygen and ionized nitrogen particles that are moving at incredible speeds due to the sun's heat. I believe this is atmospheric drag. I'm not sure if that example helps, or if this problem is too old to get any attention, but hey might as well post this.

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  • $\begingroup$ Air drag goes by V^2. I also was just thinking how an ice hockey puck going really fast takes linearly longer amount of time to slow down. Air hockey has just the drag part, which is very small at low speed. ;p $\endgroup$
    – atlex2
    Jun 22, 2019 at 22:25
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There are a number of wrong assertions in your statement. You say, "surely ... continue accelerating to infinity." Since $ a = \frac {F}{m} $, as long as you apply net force F, you only get constant acceleration a. The acceleration value not only does not go to infinity, but actually does not change, unless the net force changes. When there is no net force, the acceleration goes to zero!

What changes is the speed of the object ( $V = V0 + at$), as long as the acceleration is present.

Using the values you provide, the corresponding accelerations would be $$a1 = \frac {50- (.5)(10)(9.8)}{10} = 0.10 \text{ } m/{s^2} $$ $$a2 = \frac {60 - (.5)(10)(9.8)}{10} = 1.10 \text{ }m/{s^2}$$.

Starting from V0 = 0 and applying the 50N force for 60 seconds, the object would attain a speed of $$V1 = 0 + (.1)(60) = 6\text{ } m/s$$ If we now change the force to 60N, and apply it for 30 seconds, the new speed would be $$V2 = 6 + (1.1)(30) = 39 \text{ } m/s$$ Since the coefficient for static friction (not given) is normally higher than kinetic friction, the object most likely will not move with the 50N force. If this is the case, then the final velocity would be 33 m/s

As you may have noticed, I had to arbitrarily specify the length of time of the applied forces, otherwise an answer could not be obtained.

Note: acceleration due to gravity = 9.8 m/${s^2}$.

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