This is a question from my son’s homework, already submitted and marked.
I was a physics teacher and am curious about one of the answers.
A person pushes a block a distance of 3.0m with a force of 250N 25 degrees from the horizontal. The coefficient of kinetic friction is 0.1. How much work does the person do?
I see three possible answers.
250N x 3m
b) 250Ncos25 x 3m
c) (250Ncos25 - force of friction) x 3m
The correct answer (supposedly) is a).
I wonder why it is not b) since the vertical component of the force isn’t doing any work.
The key formula here is: work = force times displacement in that direction. The thing about your question is, it does matter if your force is directed parallel to the displacement or not (e.g., if a block is sliding across a flat floor, the weight of the block is obviously not doing any work, even though it is present as a force).
ETA never mind, ninjaed. Anyway, check the problem for an accompanying diagram. Maybe the block is being pushed straight up a ramp.
The question, at least as you’ve stated it, is ambiguous. Since “distance” is a scalar, we don’t have enough information regarding in which direction the block moved. When I first read the problem, I assumed that the force and the movement were aligned, both 25 degrees from horizontal, which makes #1 the right answer. When you read it, it seems you assumed that they were offset 25 degrees, making #2 the right answer. The question tells us the force is 25 degrees from horizontal but is silent about the block’s displacement direction.
I’m slow enough to post that I see others have responded. But since your restatement of the problem contains your unstated assumption, I’m going to post this anyway and encourage you to check whether that assumption affected your interpretation of the original problem as it was given.
The setup has defined the force exerted by the person as 250 N, so it doesn’t matter if the coefficient of friction is 0.1 or 10,000.
Mechanical work is the dot product of the force vector and the motion vector - IOW, you take magnitude of the motion vector and multiple it only by the component of the force vector that is parallel to the motion vector.
So if you’ve got 250 N of force directed 25 degrees from horizontal, and the block is moving 3 meters horizontally, then b) is the correct answer.
Physics problems often ignore that part and it’s easy to have a real life situation where you pull something with a force not parallel to the motion, so it’s a standard problem that someone has changed slightly and made “non-practical”.
The only way it’s (a) is if the block is sliding across a surface that is parallel to the force, i.e. being pushed up or down a 25 degree incline. The question is actually ambiguous on this point.
For what it’s worth, on the issue of ambiguity - before I had read further, I parsed the first sentence of the question as saying that the block was being pushed up a 25-degree slope.
I wrote the OP and post 4.
The floor is horizontal.
The push is 25 degrees up from the horizontal directed slightly down on the object.
I do this all the time if I am pushing something across the floor and I am taller than the object.
Sorry, I don’t know how to post a picture.
You might push on an object downward at an angle because the handle on the object (or whatever part you’re pushing on) is lower than your shoulder height. You might push on an object upwards at an angle in order to decrease the normal force between the object and the floor, and hence the friction. You might push at an angle in either direction because the object is in danger of tipping over, and so you want to minimize the torque.
I agree that the question is ambiguous. I had presumed that the block was being pushed up a ramp, but I have to admit that this isn’t explicit in the question. It might be a diagonal force on a horizontal floor. I think the OP should point this out to the author of the homework assignment.
Where are you getting this extra info? You need to post the actual, exact wording of the question, otherwise there’s no way to judge which answer is correct.
ETA: We all agree that, if the floor is horizontal and the force direction is 25 degrees from horizontal, then the answer is (b). What’s not clear is whether that’s the correct interpretation of the question.
Because the person pushing cannot orient their application of force directly through the center of mass parallel to the ground. In fact, unless the person pushing is able to push in such a way that their weight (due to acceleration from gravity) counteracts the couple that occurs as a result of the fact that the person is applying a force to the box at some height above the floor while the reaction force is applied to the feet at the floor, they will have to push up at some angle to keep from falling backward.
The answer to the problem is, in fact, a). The problem statement is a bit of a trick in that they give you the coefficient of friction (𝜇[SUB]k[/SUB]=0.1) but not the mass of the box, which means you can’t calculate the force necessary to overcome friction and move the box. However, you don’t need it because they give you the necessary force (F=250N at a 25º angle), which is applied for the 3 meters of movement; hence, W = F × d = 250 N × 3 m = 750 N∙m. That the force is applied at an angle is irrelevant; this is the level of force to achieve the necessary work to move the box for that distance, and of course all of it is unrecoverable because it is all lost to friction.
The mass of the box, if I have done my trigonometry and units correctly, is ~230 kg.
Stranger, I’d expect better from you. Yes, the person is exerting a 250 N force, but not all of that force is doing work. If an elephant weighing 5000 N is standing on top of the box while it’s being pushed, is the elephant doing 15,000 N of work?
OP here.
The mass of the box is given in the question. I don’t remember what it is. I don’t think it is relevant to what I am looking for.
Also, I assume that the person is pushing through the center of gravity of the box.
Sorry, I don’t understand the ambiguity in the question. It’s a pretty standard beginners physics question.
So far, most have chosen b) but one has firmly chosen a).
A person pushes a block a distance of 3.0m with a force of 250N 25 degrees from the horizontal.
Does “25 degrees from the horizontal” refer to the displacement vector of 3.0m that the block is being pushed, or does it refer to the direction of the force vector, implying that the displacement is horizontal? Grammatically it could mean either.
By the way, even if we use the latter interpretation, 25 degrees from the horizontal would mean 25 degrees above the horizontal, and your later posts seem to interpret it as 25 degrees below the horizontal, i.e., pushing downwards (which is -25 degrees from horizontal).