OCR M2 - Question on Power/Work Done (from June 2014 Question 5) Watch

KelvZhan
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Hi there,

I've encountered a difficult question regarding energy and work done, and I'm struggling to make sense of it. This is Question 5 of 4729/01 (OCR M2) June 2014, as follows:

[I have completed (i), but am struggling on (ii).]

(i) A car of mass 800kg is moving at a constant speed of 20ms-1 on a straight road down a hill inclined at an angle α to the horizontal. The engine of the car works at a constant rate of 10kW and there is a resistance to motion of 1300N. Show that sin α = 5/49. [4]

(ii) The car now travels up the same hill and its engine now works at a constant rate of 20kW. The resistance to motion remains 1300N. The car starts from rest and its speed is 8ms-1 after it has travelled a distance of 22.1m. Calculate the time taken by the car to travel this distance. [5]

The mark scheme states that part (ii) involves using Power = Work Done / Time. That part should not be a problem. My difficulty is how to calculate Work Done in the first place. In terms of my understanding, I'm also confused as to whether the weight of the car travelling up the slope is included in the resistance to motion.

Mark scheme: http://www.ocr.org.uk/Images/235978-...ics-2-june.pdf
Examiner's report: http://www.ocr.org.uk/Images/178014-...eport-june.pdf

Edit - solution found:

(Original post by Warglebargle)
Oh, I see where I went wrong. WD = Fs = (1300)(22.1) = 28730JSo redoing the lines from there:WD = Energy Gained + Energy of resistance to motion = 43280 + 28730 = 72010J.Time = WD / Power = 72010 / 20000 = 3.60sBingo. Thanks for your help!---Since I have your attention, are there any questions you have come across that use Fscos\theta instead of just Fs? This is the first past paper I have done to prepare for mocks after the half term holiday, so as a result, I'm still fuzzy on when I should use Fs and Fscos\theta.Also, I initially got (i) wrong because I thought F in P=Fv was (Driving Force - Resistance of 1300N). Instead, it turns out that F is the driving force. Do you know of any resources that could clear up this confusion?Many thanks!
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samb1234
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(Original post by Warglebargle)
Hi there,

I've encountered a difficult question regarding energy and work done, and I'm struggling to make sense of it. This is Question 5 of 4729/01 (OCR M2) June 2014, as follows:

[I have completed (i), but am struggling on (ii).]

(i) A car of mass 800kg is moving at a constant speed of 20ms-1 on a straight road down a hill inclined at an angle α to the horizontal. The engine of the car works at a constant rate of 10kW and there is a resistance to motion of 1300N. Show that sin α = 5/49. [4]

(ii) The car now travels up the same hill and its engine now works at a constant rate of 20kW. The resistance to motion remains 1300N. The car starts from rest and its speed is 8ms-1 after it has travelled a distance of 22.1m. Calculate the time taken by the car to travel this distance. [5]

The mark scheme states that part (ii) involves using Power = Work Done / Time. That part should not be a problem. My difficulty is how to calculate Work Done in the first place.

Mark scheme: http://www.ocr.org.uk/Images/235978-...ics-2-june.pdf
Examiner's report: http://www.ocr.org.uk/Images/178014-...eport-june.pdf
Consider all the different forms of energy at the beginning (when it's at the bottom) and at the end
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Zacken
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(Original post by Warglebargle)
Hi there,

I've encountered a difficult question regarding energy and work done, and I'm struggling to make sense of it. This is Question 5 of 4729/01 (OCR M2) June 2014, as follows:

[I have completed (i), but am struggling on (ii).]

(i) A car of mass 800kg is moving at a constant speed of 20ms-1 on a straight road down a hill inclined at an angle α to the horizontal. The engine of the car works at a constant rate of 10kW and there is a resistance to motion of 1300N. Show that sin α = 5/49. [4]

(ii) The car now travels up the same hill and its engine now works at a constant rate of 20kW. The resistance to motion remains 1300N. The car starts from rest and its speed is 8ms-1 after it has travelled a distance of 22.1m. Calculate the time taken by the car to travel this distance. [5]

The mark scheme states that part (ii) involves using Power = Work Done / Time. That part should not be a problem. My difficulty is how to calculate Work Done in the first place.

Mark scheme: http://www.ocr.org.uk/Images/235978-...ics-2-june.pdf
Examiner's report: http://www.ocr.org.uk/Images/178014-...eport-june.pdf
Work done = force * distance.

There is work done against both gravity and resistance to motion.
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KelvZhan
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(Original post by samb1234)
Consider all the different forms of energy at the beginning (when it's at the bottom) and at the end
Thanks for your reply.

At the beginning:
KE and PE = 0.

At the end:
KE = 1/2mv^2 = 1/2(800)(8)^2 = 25600J
PE = mgh = (800)(9.8)(22.1sinα) = (800)(9.8)(22.1*5/49)=17680J

Therefore, the energy gained = 25600 + 17680 = 43280J.

The work done = energy gained + resistance to motion = 43280 + 1300 = 44580J.

Power = work done / time, so time = work done / power = 44580/20000 = 2.229s

This is unfortunately incorrect - the time in the mark scheme is 3.60s. Where did I go wrong?
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samb1234
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(Original post by Warglebargle)
Thanks for your reply.

At the beginning:
KE and PE = 0.

At the end:
KE = 1/2mv^2 = 1/2(800)(8)^2 = 25600J
PE = mgh = (800)(9.8)(22.1sinα) = (800)(9.8)(22.1*5/49)=17680J

Therefore, the energy gained = 25600 + 17680 = 43280J.

The work done = energy gained + resistance to motion = 43280 + 1300 = 44580J.

Power = work done / time, so time = work done / power = 44580/20000 = 2.229s

This is unfortunately incorrect - the time in the mark scheme is 3.60s. Where did I go wrong?
The resistance to motion is a force, so what is the total work done against that force?
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KelvZhan
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(Original post by samb1234)
The resistance to motion is a force, so what is the total work done against that force?
Oh, I see where I went wrong. WD = Fs = (1300)(22.1) = 28730J

So redoing the lines from there:
WD = Energy Gained + Energy of resistance to motion = 43280 + 28730 = 72010J.

Time = WD / Power = 72010 / 20000 = 3.60s

Bingo. Thanks for your help!

---

Since I have your attention, are there any questions you have come across that use Fscos\theta instead of just Fs? This is the first past paper I have done to prepare for mocks after the half term holiday, so as a result, I'm still fuzzy on when I should use Fs and Fscos\theta.

Also, I initially got (i) wrong because I thought F in P=Fv was (Driving Force - Resistance of 1300N). Instead, it turns out that F is the driving force. Do you know of any resources that could clear up this confusion?

Many thanks!
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samb1234
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(Original post by Warglebargle)
Oh, I see where I went wrong. WD = Fs = (1300)(22.1) = 28730J

So redoing the lines from there:
WD = Energy Gained + Energy of resistance to motion = 43280 + 28730 = 72010J.

Time = WD / Power = 72010 / 20000 = 3.60s

Bingo. Thanks for your help!

---

Since I have your attention, are there any questions you have come across that use Fscos\theta instead of just Fs? This is the first past paper I have done to prepare for mocks after the half term holiday, so as a result, I'm still fuzzy on when I should use Fs and Fscos\theta.

Also, I initially got (i) wrong because I thought F in P=Fv was (Driving Force - Resistance of 1300N). Instead, it turns out that F is the driving force. Do you know of any resources that could clear up this confusion?

Many thanks!
It's relatively unlikely you would need fscos(x). That formula is for the case where the force the object is acting against is not parallel to the direction of motion (well technically w=fs is a special case of that general formula with theta =0 ) so unless they had some question where the object was being pulled up a ramp by a rope not parallel to the plane or something like that it wouldn't be relevant.
There's a pretty intuitive way of realising that F is the driving force. Imagine you're driving along the motorway at a constant speed of say 50ms-1. If your idea was right, then by your logic my cars engine is doing no work at all, which clearly isn't correct.
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KelvZhan
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(Original post by samb1234)
It's relatively unlikely you would need fscos(x). That formula is for the case where the force the object is acting against is not parallel to the direction of motion (well technically w=fs is a special case of that general formula with theta =0 ) so unless they had some question where the object was being pulled up a ramp by a rope not parallel to the plane or something like that it wouldn't be relevant.
There's a pretty intuitive way of realising that F is the driving force. Imagine you're driving along the motorway at a constant speed of say 50ms-1. If your idea was right, then by your logic my cars engine is doing no work at all, which clearly isn't correct.
Oh, I understand now! Thanks for the help.
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samb1234
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(Original post by Warglebargle)
Oh, I understand now! Thanks for the help.
No worries, let me know if you need any help with anything else, I took edexcel m2 last year so can probably help with most things
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KelvZhan
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(Original post by samb1234)
No worries, let me know if you need any help with anything else, I took edexcel m2 last year so can probably help with most things
I've stumbled upon another question which I'm unsure how to do. This is question 4) iii) from January 2011.

A block of mass 25 kg is dragged 30 m up a slope inclined at 5 degrees to the horizontal by a rope inclined at 20 degrees to the horizontal. The tension in the rope is 100 N and the resistance to the motion of the block is 70 N. The block is initially at rest. Calculate:

i) the work done by the tension in the rope, [2819J]
ii) the change in the potential energy of the block, [640.6J]
iii) the speed of the block after it has moved 30 m up the slope.

The mark scheme is here: http://www.ocr.org.uk/Images/65447-m...-2-january.pdf

I know that I cannot use conservation of energy where \deltaPE = \deltaKE. However, I know that there is the formula of KE(A) + PE(A) + WD(A) = KE(B) + PE(B) + WD(B), but isn't that for pulley questions? I'm not sure how to use this equation either. Could you give me pointers on how to approach this question?
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samb1234
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(Original post by Warglebargle)
I've stumbled upon another question which I'm unsure how to do. This is question 4) iii) from January 2011.

A block of mass 25 kg is dragged 30 m up a slope inclined at 5 degrees to the horizontal by a rope inclined at 20 degrees to the horizontal. The tension in the rope is 100 N and the resistance to the motion of the block is 70 N. The block is initially at rest. Calculate:

i) the work done by the tension in the rope, [2819J]
ii) the change in the potential energy of the block, [640.6J]
iii) the speed of the block after it has moved 30 m up the slope.

The mark scheme is here: http://www.ocr.org.uk/Images/65447-m...-2-january.pdf

I know that I cannot use conservation of energy where \deltaPE = \deltaKE. However, I know that there is the formula of KE(A) + PE(A) + WD(A) = KE(B) + PE(B) + WD(B), but isn't that for pulley questions? I'm not sure how to use this equation either. Could you give me pointers on how to approach this question?
You can still use conservation of energy you just need to consider all the different types of energy in the question
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KelvZhan
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(Original post by samb1234)
You can still use conservation of energy you just need to consider all the different types of energy in the question
I'm not sure how to apply the conservation of energy into this, so instead, I'm thinking that the work done by the tension provides for the gain in kinetic energy, gain in potential energy and the energy needed to overcome the resistance.

As a result, I have ended up with:
2819 = PE gained + KE gained + [Fs of resistance]
2819 = 640.6 + 1/2(25)v^2 + (30)(70)
v = 2.51ms-1

This is the correct answer. Although the method matches up with the mark scheme, is there a better way of approaching this?

Here's the confusing thing: I'm not sure when the 'resistance to motion' includes resistance due to weight on a slope. If you have time, could you take a look at June 2012's Q2) ii)?

Paper: http://www.ocr.org.uk/Images/136143-...echanics-2.pdf
Mark Scheme: http://www.ocr.org.uk/Images/135294-...ics-2-june.pdf

You see that it states the 'resistance to motion remains 800N'. In part ii, if the speed is steady, then F_{net} = 0. So F_{driving} = resistance to motion. I thought resistance to motion would be 800N, but the mark scheme throws in resistance due to weight separately!

Why is it that in the 2012 question, resistance to motion does not include weight, while in the 2011 question, resistance to motion includes weight?
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samb1234
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(Original post by Warglebargle)
I'm not sure how to apply the conservation of energy into this, so instead, I'm thinking that the work done by the tension provides for the gain in kinetic energy, gain in potential energy and the energy needed to overcome the resistance.

As a result, I have ended up with:
2819 = PE gained + KE gained + [Fs of resistance]
2819 = 640.6 + 1/2(25)v^2 + (30)(70)
v = 2.51ms-1

This is the correct answer. Although the method matches up with the mark scheme, is there a better way of approaching this?

Here's the confusing thing: I'm not sure when the 'resistance to motion' includes resistance due to weight on a slope. If you have time, could you take a look at June 2012's Q2) ii)?

Paper: http://www.ocr.org.uk/Images/136143-...echanics-2.pdf
Mark Scheme: http://www.ocr.org.uk/Images/135294-...ics-2-june.pdf

You see that it states the 'resistance to motion remains 800N'. In part ii, if the speed is steady, then F_{net} = 0. So F_{driving} = resistance to motion. I thought resistance to motion would be 800N, but the mark scheme throws in resistance due to weight separately!

Why is it that in the 2012 question, resistance to motion does not include weight, while in the 2011 question, resistance to motion includes weight?
Resistance to motion doesn't include weight. Regarding the 2011 question, if instead of thinking about the change in gpe you considered the work done against the force of gravity, you get mgsintheta x, where x is the distance along ramp it travels. You should notice that this is exactly the same as the formula for the gpe is (as it should be) so essentially by considering the gpe from an energy standpoint you are taking care of the work done against gravity
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