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-question-paper-unit-4729-mechanics-2.pdfMark Scheme:
http://www.ocr.org.uk/Images/135294-mark-scheme-unit-4729-mechanics-2-june.pdfYou see that it states the 'resistance to motion remains 800N'. In part ii, if the speed is steady, then
Fnet=0. So
Fdriving=resistancetomotion. 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?