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AQA M2 Work energy and power - not sure where to start

At last I've got to the last question in the section on energy (in pdf). It's an underlined question (q9) meaning that it's meant to be challenging. I'm unconvinced about what I should do.

So, the particle is at A and manages to stay there until it's projected up the slope so it gains in KE. (I'm assuming GPE = 0 at start of A). It goes up the slope (with no resistance from friction?) so gain in GPE = KE? Do I think about friction on the way up?

Then on the way down there must be friction as there's a mu in the answer.

I really don't know how to think this one out at all.

If someone could explain what's happening when then I might be able to have a go.

Thanks
Scan0033.pdf343.1KB
Original post by maggiehodgson
At last I've got to the last question in the section on energy (in pdf). It's an underlined question (q9) meaning that it's meant to be challenging. I'm unconvinced about what I should do.

So, the particle is at A and manages to stay there until it's projected up the slope so it gains in KE. (I'm assuming GPE = 0 at start of A). It goes up the slope (with no resistance from friction?) so gain in GPE = KE? Do I think about friction on the way up?

Then on the way down there must be friction as there's a mu in the answer.

I really don't know how to think this one out at all.

If someone could explain what's happening when then I might be able to have a go.

Thanks


On the way up, ke at start - work done against friction = pe at top

This will enable you to find the distance travelled, x. Note that x will come into the pe term too.

Then do a similar thing on the way down.
(edited 9 years ago)
Original post by tiny hobbit
On the way up, ke at start - work done against friction = pe at top

This will enable you to find the distance travelled, x. Note that x will come into the pe term too.

Then do a similar thing on the way down.



Thanks.

So I do think about friction on the way up.

It's given me a start.
Original post by maggiehodgson
At last I've got to the last question in the section on energy (in pdf). It's an underlined question (q9) meaning that it's meant to be challenging. I'm unconvinced about what I should do.

So, the particle is at A and manages to stay there until it's projected up the slope so it gains in KE. (I'm assuming GPE = 0 at start of A). It goes up the slope (with no resistance from friction?) so gain in GPE = KE? Do I think about friction on the way up?

Then on the way down there must be friction as there's a mu in the answer.

I really don't know how to think this one out at all.

If someone could explain what's happening when then I might be able to have a go.

Thanks


Define a set of co-ordinate axis which has 'x' parallel to the inclined plane and 'y' perpendicular to it.

Use F=mv⋅d(v)dx\textbf{F}= \textbf{mv} \cdot \frac{d(\textbf{v})}{d\textbf{x}} and integrate with appropriate limits. Your forces will be the weight of the particle and friction but you must define these carefully with respect to your co-ordinate axis.

Repeat for trip of the particle back down the inclined plane and add together your two values for distance traveled.
Original post by pleasedtobeatyou
Define a set of co-ordinate axis which has 'x' parallel to the inclined plane and 'y' perpendicular to it.

Use F=mv⋅d(v)dx\textbf{F}= \textbf{mv} \cdot \frac{d(\textbf{v})}{d\textbf{x}} and integrate with appropriate limits. Your forces will be the weight of the particle and friction but you must define these carefully with respect to your co-ordinate axis.

Repeat for trip of the particle back down the inclined plane and add together your two values for distance traveled.


I think this will be beyond the current knowledge of the OP
Original post by tiny hobbit
I think this will be beyond the current knowledge of the OP


My bad.
Original post by tiny hobbit
I think this will be beyond the current knowledge of the OP


You are correct. It is beyond me.
Original post by pleasedtobeatyou
My bad.



No problem. It has made me realise that I have a very long way to go and has given me something to look forward to.

Thanks for trying, anyway.
Original post by tiny hobbit
On the way up, ke at start - work done against friction = pe at top

This will enable you to find the distance travelled, x. Note that x will come into the pe term too.

Then do a similar thing on the way down.



Oh dear.

I've now been able to have a go but it looks horrible. I'm sure it won't simplify to what required as there's a sinTheta in the numerator!

Here's my workings. Pointers please.
Scan0001.pdf316.3KB
Original post by maggiehodgson
Oh dear.

I've now been able to have a go but it looks horrible. I'm sure it won't simplify to what required as there's a sinTheta in the numerator!

Here's my workings. Pointers please.


Not checked all of it, but I note that when working out the work done against friction, you took the normal reaction to be mcos⁡θm\cos\theta, rather than mgcos⁡θmg\cos\theta

Edit: Also, for the second part C to B, your loss in GPE is incorrect, since you've used the GPE at C taking A as zero. Doing that, the GPE at B is not zero (relative to A), so the loss in GPE is not simply the GPE at C.

A slightly simpler way for CB in the spoiler - use, or not, as you wish.

Spoiler

(edited 9 years ago)

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