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M2 Question - Work, Energy and Power

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1. Hi all,

I'm stuck on a question that seems to make no sense:

"A barge is pulled along a straight canal by a horse on the towpath. The barge and the horse move in parallel straight lines 5m apart. The tow-rope is 13m long and it remains taut and horizontal. The horse and the barge each move at a constant speed of 0.78ms-1 and the tow-rope has a constant tension of 400N. Calculate the work done by the horse on the barge in 10 minutes."

I just can't make any sense of this; if the barge and horse are 5m apart, then how can the tow-rope between them be 13m and taut?

Anyway, I had a crack at the question:

s = v*t = 0.78 * 600 = 468m

W.D = Fs = 400 * 468 = 187kJ (3sf)

The answer was 173kJ, and I have no idea how they got that. Any help would really be appreciated :)
2. (Original post by confusedmathser)
Hi all,

I'm stuck on a question that seems to make no sense:

"A barge is pulled along a straight canal by a horse on the towpath. The barge and the horse move in parallel straight lines 5m apart. The tow-rope is 13m long and it remains taut and horizontal. The horse and the barge each move at a constant speed of 0.78ms-1 and the tow-rope has a constant tension of 400N. Calculate the work done by the horse on the barge in 10 minutes."

I just can't make any sense of this; if the barge and horse are 5m apart, then how can the tow-rope between them be 13m and taut?

Anyway, I had a crack at the question:

s = v*t = 0.78 * 600 = 468m

W.D = Fs = 400 * 468 = 187kJ (3sf)

The answer was 173kJ, and I have no idea how they got that. Any help would really be appreciated
Imagine a triangle, the vertices connecting two ends of the hypoteneuse are the horse and barge and this is13 metres. The base of the triangle is 5 metres and the height (you can find) is 12 metres. When it says they are 5 metres it means that the horse is moving in a straight line and so is the barge; the distae between these straight lines is 5 metres but the horse is further ahead, obviouly.

Apologies for the typo, on my phone.
3. (Original post by confusedmathser)
Hi all,

I'm stuck on a question that seems to make no sense:

"A barge is pulled along a straight canal by a horse on the towpath. The barge and the horse move in parallel straight lines 5m apart. The tow-rope is 13m long and it remains taut and horizontal. The horse and the barge each move at a constant speed of 0.78ms-1 and the tow-rope has a constant tension of 400N. Calculate the work done by the horse on the barge in 10 minutes."

I just can't make any sense of this; if the barge and horse are 5m apart, then how can the tow-rope between them be 13m and taut?

Anyway, I had a crack at the question:

s = v*t = 0.78 * 600 = 468m

W.D = Fs = 400 * 468 = 187kJ (3sf)

The answer was 173kJ, and I have no idea how they got that. Any help would really be appreciated
You'll need to think about resolving here, the barge has what resultant force acting forward? What do you know about Force, velocity and power? Powerr = Force * velocity and power = work done/time

Remember that constant speed = no acceleration = no resultsnt force. So that'll let you set up an equation for the foce on the barge and solve for work done.
4. Perhaps worth pointing out that in the formula,

W.D = Fs

the s represents the distance moved in the direction of that force.

Hence as Zacken pointed out, the need to resolve. There is no work done by the force perpendiuclar to the motion of the barge, as there is no movement in that direction. The only work is done by the component of the force in the direction the barge is moving.
5. (Original post by Zacken)
Imagine a triangle, the vertices connecting two ends of the hypoteneuse are the horse and barge and this is13 metres. The base of the triangle is 5 metres and the height (you can find) is 12 metres. When it says they are 5 metres it means that the horse is moving in a straight line and so is the barge; the distae between these straight lines is 5 metres but the horse is further ahead, obviouly.

Apologies for the typo, on my phone.
Thanks, took me a few minutes to get my head around that even with your explaination, but now I've done a sketch I get it. I've got the right answer now too. Thanks for the help, I just hope they don't word a question like that in the exam
6. (Original post by confusedmathser)
Thanks, took me a few minutes to get my head around that even with your explaination, but now I've done a sketch I get it. I've got the right answer now too. Thanks for the help, I just hope they don't word a question like that in the exam
I can almost guarantee that if you are doing Edexcel you will be asked about trucks on inclined planes instead.

Well done on getting the answer!
7. (Original post by ghostwalker)
Perhaps worth pointing out that in the formula,

W.D = Fs

the s represents the distance moved in the direction of that force.

Hence as Zacken pointed out, the need to resolve. There is no work done by the force perpendiuclar to the motion of the barge, as there is no movement in that direction. The only work is done by the component of the force in the direction the barge is moving.
Thanks, once I understood what the question was trying to tell me I got the answer Does the question seem really oddly worded to you or is it just me?
8. (Original post by Zacken)
I can almost guarantee that if you are doing Edexcel you will be asked about trucks on inclined planes instead.

Well done on getting the answer!
I'm on OCR mate, really hope they ask the inclined planes question as well because those are so much easier imo
9. (Original post by confusedmathser)
I'm on OCR mate, really hope they ask the inclined planes question as well because those are so much easier imo
Haha, yeah. Fair play. I'm sure you'll be fine! :-)
10. (Original post by confusedmathser)
Thanks, once I understood what the question was trying to tell me I got the answer Does the question seem really oddly worded to you or is it just me?
Wording seems fine to me.

But I've watched enough programmes on canals to know what the arrangement is.

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