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# Edexcel M2 Work-Energy Mock paper QUESTION watch

1. Hi there, got a question from a M2 mock paper that i'm having some "doubts" about, so here to share it to see if anyone can enlighten me somewhat

Qu. 8 (a)
A partile P is projected up a line of greatest slope of a rough plane which is inclined at an angle alpha to the horizontal, where tan alpha=3/4
The coefficient of friction between P and the plane is 1/2
The particle is projected from the point O with a speed of 10m/s and
comes to instantaneous rest at the point A.

Using Work-Energy principle, or otherwise,

(a) find, to 3significant figures, the length of OA.

My method:
GAIN IN GPE = LOST IN KE

Letting x = OA (the length of OA)
x multiplied by sin alpha must be the VERTICAL height - right?

mg * xsin alpha = 1/2(u^2 - v^2)*m

mg * xsin alpha = 1/2(10^2 - 0^2)*m

As tan alpha = 3/4, therefore sin alpha = 3/5

m's CANCEL out, sin alpha replaced with 3/5

3gx = 50

x = 8.5m = OA

Which does not correspond with the mark scheme answer. I'm wondering... does it differ because there is a ROUGH plane?

This is highly confusing me because i've always been taught this method, and somehow out pops this that totally contradicts with what i've been taught.

Much appreciated with anyone can guide this in the right direction
2. (Original post by LT0918)
Hi there, got a question from a M2 mock paper that i'm having some "doubts" about, so here to share it to see if anyone can enlighten me somewhat

Qu. 8 (a)
A partile P is projected up a line of greatest slope of a rough plane which is inclined at an angle alpha to the horizontal, where tan alpha=3/4
The coefficient of friction between P and the plane is 1/2
The particle is projected from the point O with a speed of 10m/s and
comes to instantaneous rest at the point A.

Using Work-Energy principle, or otherwise,

(a) find, to 3significant figures, the length of OA.

My method:
GAIN IN GPE = LOST IN KE

Letting x = OA (the length of OA)
x multiplied by sin alpha must be the VERTICAL height - right?

mg * xsin alpha = 1/2(u^2 - v^2)*m

mg * xsin alpha = 1/2(10^2 - 0^2)*m

As tan alpha = 3/4, therefore sin alpha = 3/5

m's CANCEL out, sin alpha replaced with 3/5

3gx = 50

x = 8.5m = OA

Which does not correspond with the mark scheme answer. I'm wondering... does it differ because there is a ROUGH plane?

This is highly confusing me because i've always been taught this method, and somehow out pops this that totally contradicts with what i've been taught.

Much appreciated with anyone can guide this in the right direction
The frictional force F is the normal reaction(N) * µ

Resolve perp. to slope:
N = (4mg)/5 so F= (2mg)/5

So work done against friction is Fx = (2mgx)/5

PE gained + work done against friction = KE lost

(3xmg)/5 + (2xmg)/5 = (m/2)*100

xmg = 50m
xg = 50
x=50/9.8 = 5.1

Aitch
3. OK, thank you

So basically PE GAINED = KE LOST has to take frictional work done into account when there is a rough plane.

But lets just say that OA was 8.5m and pretend that it is the distance OA.

And lets imagine we're now trying to find the vertical height it moved up.

So we have: Sin alpha = Vertical / 8.5

3/5 = Vertical / 8.5

Vertical = 5.1m

Can this be explained ?

Much appreciated. I'm accepting the fact that you have to take friction into account, but would be grateful if anyone can also explain the above.

Cheers
4. (Original post by LT0918)
OK, thank you

So basically PE GAINED = KE LOST has to take frictional work done into account when there is a rough plane.

But lets just say that OA was 8.5m and pretend that it is the distance OA.

And lets imagine we're now trying to find the vertical height it moved up.

So we have: Sin alpha = Vertical / 8.5

3/5 = Vertical / 8.5

Vertical = 5.1m

Can this be explained ?

Much appreciated. I'm accepting the fact that you have to take friction into account, but would be grateful if anyone can also explain the above.

Cheers

Roughly

KE lost = 50m

It becomes Work Done = 20m

Plus PE gained = 30m

If you omit the work done and equate KE lost and PE gained (say you oil the slope!) The work done against friction will all become PE, so the height gained will increase proportionately. This proportion is (see above) about (5/3)

5.1*(5/3) = 8.5

Aitch
5. (Original post by Aitch)
Roughly

KE lost = 50m

It becomes Work Done = 20m

Plus PE gained = 30m

If you omit the work done and equate KE lost and PE gained (say you oil the slope!) The work done against friction will all become PE, so the height gained will increase proportionately. This proportion is (see above) about (5/3)

5.1*(5/3) = 8.5

Aitch
(PE gained)/(KE lost) just happens to equal about sin alpha!

It wouldn't if (for example) u had a different value!

Aitch
6. Thank you very much, I've accepted the fact now and also it just "happens" to work with other similiar questions too :P

Much appreciated =)

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