Shouldn’t the slopped distance be considered instead of the height?
(Sayonara Zetsubou Sensei: chapter 233)
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/s2 * 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.
work = average force • displacement = |average force| * |displacement| * cos(angle between them) = component of average force parallel to displacement * |displacement| = component of displacement parallel to force * |average force| weight = force due to gravity = mass * acceleration due to gravity ≈ 5 kg * 9.8 m/s² = 49 N work = component of displacement parallel to average force * |average force| = 2 m * weight ≈ 98 NmOP specifically asked about the sloped distance:
Shouldn’t the slopped distance be considered instead of the height?
Ah, my mistake.
