# I Confusion about work performed by a horizontal push or pull

1. Jun 13, 2017

### NekotoKoara

Hello,

Over the past couple of weeks I have been working my way through the basics of physics on my own. I've just arrived at work and energy and there is one thing that confuses me a little bit. Let's say I have a 25kg box on a sidewalk. I go to push it and I ultimately push it 5m east (lets say east is the positive direction in this example) over a period of 5 seconds. Normally I see physics problems already include a specific force acted on the box by the person pushing it. For example they will say Sally pushed a box 5 meters with a force of 20 newtons. Therefore Sally exerted 100 joules of work. But what if the force exerted is unknown? How would you derive the force from the information I provided (25kg box, +5m, 5 seconds)? Is it even possible to derive from that information? If not, what other information would be required to figure out the force that was applied?

The biggest point of confusion for me is the acceleration. If this were discussing an object dropping down with gravity acting on it then it seems much more straight forward since gravity, for the most part, is pretty consistent at Earth's surface. However a horizontal push, in real life, could be much more inconsistent. The box could start at 0m/s then accelerate up to 1m/s over a period of 1 second, have a constant velocity for 2 seconds, then over the rest of the journey accelerate back down to 0m/s. Or it could have any other of an infinite amount of velocities over an infinite amount of periods of time during that 5 second period. Is there some fundamental concept that I am completely missing or misunderstanding that will help alleviate my confusion?

Sorry if this is at all unclear. Any sort of help clearing this all up for me would be greatly appreciated. :-)

2. Jun 13, 2017

### Dr.D

If you only know the mass of the box, the distance travelled, and the total time, you really cannot determine the force involved in pushing the box. For example, the force might be just enough to make it move (to overcome friction), for the first 4.9 seconds, but then an extremely large value to complete the motion in the last 1/10 s. But there are other possibilities as well. You need to know more about the system to answer your question.

3. Jun 13, 2017

### NekotoKoara

Thank you for your response. That is what I was figuring. I'm just struggling to find a way where work could be figured out without the help of a machine telling me the force that is being exerted on an object. For example, provided I know things like the the coefficients for both static and kinetic friction, the mass of the object, etc, lets say I have an object at rest. I start pushing it and accelerate it up to a specific velocity over a specific amount of time. I then I keep pushing it at that same velocity for another specific amount of time. I then start to accelerate it down until it reaches a velocity of 0m/s. And that whole process moves the object a specific distance. Would the net force applied on the object during that process be the sum of the magnitudes of accelerations experienced both at the beginning and the end of that process multiplied by the mass of the object? If not is there some other way to use the info provided to figure out the net force applied during this process?

Again sorry if this isn't the clearest explanation.

4. Jun 13, 2017

### Dr.D

You describe a process in which you apply a force sufficient to overcome static friction, then accelerate the object up to a particular speed, hold it at that speed, and then allow it to slow to a stop. Your question,

implies that you do not understand the term "net force." The net force is the total force acting on the object at any instant. For the process you have described, the net force varies from instant to instant. In the beginning of your scenario, the net force is zero, with the applied force just approaching the friction limit. Then, as the object accelerates, the net force is the vector sum of the externally applied force and the friction force. While the object is moving at constant speed, the net force is zero, and so on...

5. Jun 13, 2017

### SlowThinker

The good news is that you can ignore acceleration.
1. Imagine there is no friction. Then you can apply a really really small force for a short time to get the box moving, wait some time until it moves the 5 meters, and stop it. Very little work is done to move it.
2. The part where you stop the box, it actually gives energy back, so the work to move the box, if there is no friction, is in fact zero.
3. You can push the box really hard or softly or even backwards. Whenever you stop it, you're back at zero energy. So the acceleration profile isn't at all important in this case.
4. If there is friction, you can split the force at any moment into a part that is the same as in the frictionless case, and a part that overcomes the friction. The first part does no work in the end, so you only need to care about the second part.
5. The force needed to overcome friction depends on the weight of the box, and the friction coefficient. If you know the two, you can find the energy needed to move the box.

6. Jun 13, 2017

### Staff: Mentor

Let's take a problem and break it into 3 parts. For the 1st part, let's assume the force applied to the box is 100 N and the opposing force of friction is 50 N, for a net force of 50 N. The box has a mass of 25 kg.

During the 1st part of the problem, where you're accelerating the box:
Once static friction is broken the acceleration is: $a=\frac{f_{net}}{m}=\frac{50}{25}=2 ms^{-2}$
Displacement required to accelerate to 5 m/s: $d=\frac{v_f^2-v_i^2}{2a}=\frac{25}{4}=6.25 m$
Total work performed by the accelerating force: $W=fd=100*6.25=625 J$

The total work can be thought of as being divided between accelerating the box and counteracting the force of friction.
Work done to accelerate the box is the same as the kinetic energy: $W=ke=0.5mv^2=0.5*25*25=312.5 J$
Another way of finding the work done to accelerate the box is to use the net force: $W=f_{net}d=50*6.25=312.5 J$
Work done against friction is the remaining force (applied force minus net force) times the displacement: $W=fd=50*6.25=312.5 J$

The work done to accelerate the box plus the work done against friction is the same as the total work.

For the 2nd part, let's assume that whoever is pushing the box continues to push the box for 10 meters at a steady velocity. Since the box isn't accelerating, the magnitude of the applied force is the same as the opposing force of friction.
Total work done: $W=fd=50*10=500 J$

For the third part, the person stops pushing the box and it slides to a halt. The applied force by the person is zero and the force of friction is still -50 N (force is a vector and the force of friction points opposite to the direction of motion). By symmetry we know that the displacement of the box during this period is the same as the displacement during the initial acceleration of the box since the magnitude of the net force is the same.
Work done is negative: $W=fd=-50*6.25=-312.5 J$
This makes sense because the work done is solely to counteract the motion of the box and must be equal (and opposite) to the box's kinetic energy.

Notice that the person only applies a force during parts 1 and 2, not 3.
The total work done by the person on the box is the sum of the calculations earlier, or: $W=f_1d_1 + f_2d_2 = 100*6.25 + 50*10 = 1135 J$

For simple problems aimed at students taking physics for the first time, we often assume that the person pushing the box only accelerates it to a negligible velocity. The removes parts 1 and 3 from the problem and lets us talk about the work done solely to move the box a certain distance against friction, greatly simplifying things.