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M2 Inclined Plane Problem

I don't understand why sometimes the weight component down the slope is ignored when calculating resistance and using it in things such as work done.
For example in question 5 part iii on June 2007 they ignore the weight, however in question 3 on June 2008 they take into account the weight acting down the slope. I suspect it is something to do with the constant resistive force given but it still remains a little unclear and I was wondering if anyone could shed some light on it.

Reply 1

Zacken

Original post by Gabenite

They have already given you the work done in moving, so the weight has been taken into account.

however in question 3 on June 2008 they take into account the weight acting down the slope. I suspect it is something to do with the constant resistive force given but it still remains a little unclear and I was wondering if anyone could shed some light on it.

Here they haven't given you the work done, so the weight hasn't been taken into account.

Reply 2

Gabenite

Original post by Zacken

They have already given you the work done in moving, so the weight has been taken into account.

Here they haven't given you the work done, so the weight hasn't been taken into account.

that makes more sense but I'm still confused, in the question where work done is given you use change in energy = 8000 - 60*distance but I don't see why it isn't 8000 - (60 + weight down slope)*distance because surely they are both acting against it, is it possible the weight is counted within the 60N resistive force or have I completely lost the plot?

Reply 3

Zacken

Original post by Gabenite

You need to do:

work done in moving from A to B (this includes the weight) = increase in KE + increase in GPE + work done against resistance force.

Reply 4

Gabenite

Original post by Zacken

You need to do:

work done in moving from A to B (this includes the weight) = increase in KE + increase in GPE + work done against resistance force.

but why isn't work done against the weight like it is done against the resistive force?

edit: do you mean the 8000 = work done moving from A to B - work done against weight. If this is the case how am I to tell in the question whether to ignore the weight or not?

(edited 8 years ago)

Reply 5

Zacken

Original post by Gabenite

Because the word done against weight has already been given to you as 8000.

Reply 6

Gabenite

Original post by Zacken

Because the word done against weight has already been given to you as 8000.

but I thought the 8000 was the driving force * the distance up the plane so I don't see how it includes the weight?

Reply 7

Zacken

Original post by Gabenite

but I thought the 8000 was the driving force * the distance up the plane so I don't see how it includes the weight?

That doesn't make sense. Force has units newtons, 8000 is the work done and has units joules. How could it be the driving force?

Reply 8

Gabenite

Original post by Zacken

That doesn't make sense. Force has units newtons, 8000 is the work done and has units joules. How could it be the driving force?

Well yes that's why I said it was the driving force * distance up the slope

Reply 9

Zacken

Original post by Gabenite

Well yes that's why I said it was the driving force * distance up the slope

Oh sorry, my bad. It certainly includes that, but it also includes the weight. It's work done from moving from A to B and that includes both the driving force and the weight.

So it's like saying you'd re-use driving force again in your equation, that doesn't make sense because it's already been accounted for in the 8000; just as weight has been.

Reply 10

Gabenite

Original post by Zacken

But in that case why has the 60N not been accounted for the in 8000?

Reply 11

Zacken

Original post by Gabenite

But in that case why has the 60N not been accounted for the in 8000?

Because that's due to the constant resistive force.

Reply 12

Gabenite

Original post by Zacken

Because that's due to the constant resistive force.

But the work done moving from A to B must include this resistive force since work has to be done against it to move from A to B?

Reply 13

tiny hobbit

Original post by Gabenite

Firstly the June2007 question is about work and energy, whereas the June 2008 question is about power and therefore uses F = ma. In the latter case, all of the forces must be included in f = ma,so that includes the weight.

Now to the "they've ignored the weight" in the 2007 question. "Gravitational potential energy" is another name for the "work done against the weight" so you must only include one of these, otherwise you are including the same term twice.

(edited 8 years ago)

Reply 14

Gabenite

Original post by tiny hobbit

Firstly the June2007 question is about work and energy, whereas the June 2008 question is about power and therefore uses F = ma. In the latter case, all of the forces must be included in f = ma,so that includes the weight.

Now to the "they've ignored the weight" in the 2007 question. Gravitational potential energy" is another name for the "work done against the weight" so you must onlt include one of these, otherwise you are includng the same term twice.