Gravitational potential energy equal to work done against resistances ?

Greetings,

I am currently going through some past papers for M2 and I'm a bit confused about a solution suggested by the examiners. It's Question 3, part iii) shown below. I've also posted a link to the actual paper and mark scheme.

http://mei.org.uk/files/papers/m206ju_rw67.pdf

I don't get the part where they say "Extra GPE balances work done against resistances." The way I understand the following equation is:

mgxsina = GPE at point P
(x+3) is the distance travelled by the object on the way up and all the way down.

I just don't see why GPE at that point would be equal to the work done against resistances and why the distance would be expressed as (x+3).

What is it that I'm missing about the relationship between the potential energy and work done?

Thank you in advance for your time and assistance.

(edited 5 years ago)

mgxsinα is the difference in GPE between P and Q (obviously higher GPE is at P).

The examiners' solution is that the difference in total energy is equal to the work done to overcome resistance forces (without resistance, the total energy would be conserved). So the change in total energy = work done on resistance forces = 6(x+3) + 0.2 X 11g X cosα (x+3)
Since the initial speed and the speed at Q are the same, there is no change in kinetic energy.
So the change in total energy = change in potential energy = mgxsinα

Reply 2

Sinnoh

22

Conservation of energy yo. It's a closed system.

Reply 3

DeadManProp

OP

10

Original post by JSG29

mgxsinα is the difference in GPE between P and Q (obviously higher GPE is at P).

The examiners' solution is that the difference in total energy is equal to the work done to overcome resistance forces (without resistance, the total energy would be conserved). So the change in total energy = work done on resistance forces = 6(x+3) + 0.2 X 11g X cosα (x+3)
Since the initial speed and the speed at Q are the same, there is no change in kinetic energy.
So the change in total energy = change in potential energy = mgxsinα

Oh wow, I've had to chew on this for a moment, but your explanation has helped a lot. Just to double check. If the speed of the object has actually changed, would the equation then be: [change in GPE] + [change in KE] = [work done against resistances] ?

Thank you once again!

Reply 4

JSG29

11

Original post by DeadManProp

Oh wow, I've had to chew on this for a moment, but your explanation has helped a lot. Just to double check. If the speed of the object has actually changed, would the equation then be: [change in GPE] + [change in KE] = [work done against resistances] ?

Thank you once again!

Yes - just as a quick sidenote that I forgot to mention earlier, it should be [decrease in GPE] + [decrease in KE] = [work done against resistance] as work done is removing energy from the system.

Reply 5

DeadManProp

OP

10

Original post by JSG29

Yes - just as a quick sidenote that I forgot to mention earlier, it should be [decrease in GPE] + [decrease in KE] = [work done against resistance] as work done is removing energy from the system.

Hmm, what if an object is still moving up a slope? Would that be an increase in GPE? And what would the equation look like if a driving force was introduced and the object was accelerating up a slope? [Increase in GPE] + [Increase in KE] = [Work done by driving force] - [Work against resistances] ?

Sorry to pester you like this, but these papers include a lot of these sorts of questions, so I want to make sure I'm thinking about these scenarios in the right way.

Thank you. :]

Reply 6

JSG29

11

Original post by DeadManProp

Hmm, what if an object is still moving up a slope? Would that be an increase in GPE? And what would the equation look like if a driving force was introduced and the object was accelerating up a slope? [Increase in GPE] + [Increase in KE] = [Work done by driving force] - [Work against resistances] ?

Sorry to pester you like this, but these papers include a lot of these sorts of questions, so I want to make sure I'm thinking about these scenarios in the right way.

Thank you. :]

No worries, you seem to broadly have it correct but I'll just clarify each point as much as I can.

1. If the object was still moving up the slope, its GPE would be increasing - this could be written as a negative decrease in GPE; the equation could be written as [decrease in KE] - [increase in GPE]

2. Yes, that's exactly correct

Reply 7

DeadManProp

OP

10

Sorry, just one more thing. In point 1, shouldn't that be "[decrease in KE] - [decrease in GPE]" rather than "- [increase in GPE]" if we're talking about a "negative decrease in GPE"? Just wondering if it's a typo or if I'm misunderstandng your point.

Thank you.

(edited 5 years ago)

Reply 8

JSG29

11

No - it would remain [decrease in KE] + [decrease in GPE], with the value of the decrease in GPE being negative. I.e. If an object moved up a slope in a way that it gained 10J in GPE, the equation would be [decrease in KE] - 10 = [work done]. This is the same as [decrease in KE] + (-10)