Is this physics correct?
Shouldn't the slopped distance be considered instead of the height?
(Sayonara Zetsubou Sensei: chapter 233)
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View original on thelemmy.clubShouldn't the slopped distance be considered instead of the height?
(Sayonara Zetsubou Sensei: chapter 233)
7 replies
Hello, Physics teacher here! Key to the concept of energy aka work is its definition:
Work done on an object = force X distance moved in the direction of the force. This can also apply when the force is in the opposite direction to the movement, i.e. gravitational force down, movement up. We say we're doing work against gravity.
So the slope introduces horizontal movement but the absence of horizontal forces makes this irrelevant to the problem here. It only becomes relevant when say, sliding and friction is introduced!
No, work done only cares about the start and the end. What happened in-between doesn't matter.
In this case, the kinetic energy is 0 at the start and end, but the potential energy of the mass increased by mgh. 5 kg * 9.81 m/s^2^ * 2 m = 98.1 J (1 J = 1 N•m).
It's not as easy as applying work = force * sloped distance, since she's not simply accelerating the mass. If she did, it would have horizontal kinetic energy at the end. She's stopping it too (well, the friction is stopping it, but we're ignoring friction).
Your equations are correct (although as I said in my other comment 9.81 is the global average but gravity varies by ~0.1 so that's too many significant figures), but their issue doesn't have to do with using sloped distance, by which I assume you mean the length of the slope (they don't as far as I can tell). It says the height change is 2 meters, and they use 2 meters as the distance since that's the component of the displacement parallel to gravity. The problem is that they didn't convert mass to weight.
OP specifically asked about the sloped distance:
Ah, my mistake.
It's a deceptive thing, that incline.
As a human doing the work (human definition), it's intuitively less work to use the slope.
But in measurable, calculable physics terms, the slope only changes the time over which the Work (Physics definition) is done.
In this scenario since acceleration is near zero, it doesn't matter.
Not quite, you have to multiply distance by the force applied to the object (its weight in Newtons), but they're using its mass (kilograms).
It's actually
98 Nm, sinceweight = mass * gravitational accelerationandg ≈ 9.8 m/s²(you'll see people give more digits but it actually varies by at most0.1 m/s²or so depending on where you are).