It’s B. Think about the extreme angle cases: at 90° you’re pushing straight down, and at 0° the force you are applying is perfectly aligned with the displacement.
I’ll go you one further: I initially assumed that it referred only to the force vector, but I still don’t see how that implies anything about the direction of the displacement. I saw two interpretations of that statement:
[ul]
[li]A person pushes a block a distance of 3.0m with a force of 250N, both the same 25 degrees from the horizontal. [/li][li]A person pushes a block a distance of 3.0m, with a force of 250N @ 25 degrees measured from the horizontal. The block moved 3 meters, but we don’t know in which direction because we haven’t been told. It could have been at zero degrees along the horizontal, 25 degrees parallel to the force vector, or even at some other angle, say 15.32 degrees, because if you don’t specify the direction, I’m free to pick whatever I like. I will give you that “push” implies that the angle is no greater than 90 degrees and that the movement and force vectors aren’t skew.[/li][/ul]
I was going to point out that it is actually doing work because the upward component of the applied force is counteracted by the increased load at the front edge of the box to counter that component, but then I realized that this just reduces the static normal force of the weight of the box because the total upward reaction force is just equal to the weight regardless of the torque applied to the box from the vertical component. So you are correct, the work done is distance times the horizontal component of applied force. That also means that calculating the mass of the box has to account for that additional reaction force component, which should make it ~242 kg (assuming the box is symmetric about the center of mass.
Stranger
We have no way of knowing the mass of the box from the given information.
I think there’s a fair chance this is just a carelessly worded attempt to describe pushing a weight up an inclined plane.
If so, the correct answer is indeed a) - or, if we go by the way it’s listed in the OP, 1). And my back-of-the envelope calculation says the mass of the box is ~91 kg.
Note that this sidesteps the apparent ambiguity in “25 degrees from the horizontal”: If the box is being pushed, it must be headed up the inclined plane - with a 0.1 coefficient of kinetic friction, it would on its own slide down a 25-degree slope.
OP has still not clarified the exact wording of the original homework question, and what parts of what he has written here are his own interpretation.
Sure but no handles were mentioned, so all we can comfortably do is push the flat face with a force that’s normal to the surface.
Or, I guess if the surface is sticky or jagged, then you might get enough grip to apply some force up or down, but it’s debatable whether that should be considered “pushing”.
Personally I think this would be poor form for a beginner physics question. I suspect the original problem likely says the block is on an incline and the force is parallel to the ground.
I’ve assumed we’re pushing a face, which means our push can only be normal to the surface, and any up or down force is just acting to slide our hands up or down the face.
It seems to me to be elementary physics with a bit of diversion thrown in. Jedi mind tricks.
Work=Force x Distance
The angle of the motion, the result a potential energy stored up, the amount of energy converted to heat from friction, etc. etc. etc. - all irrelevant.
The weight was pushed 3m and it took 250N force to do it / while pushing it over that distance. work= force x distance. The answer is “1)”
The OP repeatedly insists the box was not pushed in the direction of the force. Any normal force will be cancelled out by the constraint of the floor supporting the box, which will exert a reaction force in the opposite direction. Leaving less than 250 N to actually slide the box.
This is why I asked if there was an accompanying diagram, which would make things instantly clear to the reader.
An unwarranted assumption. a person can wrap their hands around an edge formed by two perpendicular faces, allowing the push to assume pretty much any direction within a 90-degree range without relying on friction.
I just did this a few days ago. I had a new lawn mower delivered. It was packed up in a cardboard box, and after the delivery person left it just outside my garage, I went out and pushed it into my garage. Fingers on top of box, heels of palms on side of box, and push. Box was only two feet tall, so my push was definitely not oriented in the direction of travel - it might have even been pointed about 25 degrees downward. The box was 70 pounds, so it’s entirely possible the force I was exerting was somewhere in the neighborhood of 250 N.
In the first draft of my previous post, I discussed some of the complexities that may be introduced if the pusher is grabbing edges or corners. About turning forces, and how the coefficient of friction starts to actually matter etc.
But TL;DR I doubt a supposed beginner physics question is going to go there.
I still expect the original question to have a diagram where an arrow is shown pointing at a box, and the arrow is normal to the surface of the nearest face.
Showing an arrow pointing at a face but inclined relative to the face, already would make it odd for a simple mechanics question.
What’s odd about it? It doesn’t make the problem substantially more complicated (beyond the introduction of the angle, but that was probably part of the point of the problem), and it’s a fully realistic situation that occurs frequently in the real world.
Well you said it yourself with the suggestion that the block has a handle. When we’re having to imagine extra elements like that for it to make any physical sense, then it’s not a good formulation of a problem IMO.
And OK, you said a handle “or whatever”, but the point is, it’s a bit of a dirty trick that the simplest conception of “block” we can probably think of – a cube with smooth sides – doesn’t work.
Whatever it is, it has a point where we are exerting a force. In the real world, that’s likely to be a handle, but even if it isn’t, it doesn’t change anything in the problem.
You can absolutely apply a force at an angle to a cube without a handle. Even if you did need a handle to apply a force at an angle, physics homework problems are always simplified. I would not expect the diagram to show everything. It should show everything relevant and maybe some irrelevant things to see if the student can discern the relevant ones. Leaving out a handle is totally fine. This point of contention is odd to me.
If the person is pushing down, the force vector is declined 25° from the horizontal.
The physics of this problem is very straightforward but the English is lacking. The contributors here still can’t even agree on what the problem is.
Trivial if you push on an edge. If you push on a face, the friction must be sufficient to prevent sliding down the face.
Simplification is one thing, but leaving out things that are necessary for the maths to make any sense is another.
It’s not necessary. Not even a little. Put your finger on the desk in front of you and push at an angle. This is blowing your mind?
Assume we know the angle. This is completely incomprehensible unless I draw a handle on the box?