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no you are absolutely wrong,you must have made a mistake I got 2 different values.And certainly weight component has to be taken into account!! then what is the use of including hill(inclined plain) in this question??
no brother you are mistaken
hehehe I have checked it and now let me burst your bubble this question has nothing to do with OCT/NOV 2012 Qs.6i. the question you are talking about is no way linked to it.We were to find out velocities at the top of the hill and at the bottom for which we had to find out the driving force for which it is necessary to take weight component because it was going up the hill.Whereas the question you have linked t it has nothing like that we are already given velocities at bottom and hill.It just asks "what is the work done by the driving force?" a 2 mark question and as easy as abc W.D by driving force=G.P.E+W.D against resistance-loss in K.E that is it. While if you had seen Qs. 6 ii of our paper it was also asked to find out the W.D by driving force which was also unlike the question you are referring to because there was no loss of K.E there was gain of K.E and that was W.D by driving force=G.P.E+W.D against resistance+ Gain in K.E.Sorry to burst your bubble but the weight component shouldn't be taken. If you check may/june 2006 question 6 (iii), its pretty much the same math and only the resitive work done is taken.
u r probably rightman simple as that. use the formula f=ma to calculate force. subtract it with 1000 N and the substitute the values in p=fv to calculate velocities at bottom and top. use them to calculate gain in kinetic energy. Hows that?
answer was 0.9yup i got the same and wht abt last ques last part?
P3 it is!next ???? p3 ??
Oh lol thanksLet me clarify this to you.
Paper 1 - 75 marks - 30 %
Paper 3 - 75 marks - 30%
Paper 4 - 50 marks - 20%
Paper 6 - 50 marks - 20%
no no !!! I donot think ?? it contributes 30%
P1 30%
p3 30%
s1 20%
m1 = 20% m I w rong ??
hehehe I have checked it and now let me burst your bubble this question has nothing to do with OCT/NOV 2012 Qs.6i. the question you are talking about is no way linked to it.We were to find out velocities at the top of the hill and at the bottom for which we had to find out the driving force for which it is necessary to take weight component because it was going up the hill.Whereas the question you have linked t it has nothing like that we are already given velocities at bottom and hill.It just asks "what is the work done by the driving force?" a 2 mark question and as easy as abc W.D by driving force=G.P.E+W.D against resistance-loss in K.E that is it. While if you had seen Qs. 6 ii of our paper it was also asked to find out the W.D by driving force which was also unlike the question you are referring to because there was no loss of K.E there was gain of K.E and that was W.D by driving force=G.P.E+W.D against resistance+ Gain in K.E.
Hope this will clarify your doubt.I dont wanna have a fight on a question,I very well know the correct way and I am not sticking to it because I have done that way but because it a basic MECHANICS concept and a clear fact.SO lets just wait for the MS
yes I can sense the evil of itP3 it is!
hahaha no problemohh i thought you were saying that the equation should be W.D by DF=Change in M.E. + W.D. by R + W.D. by weight
how cum? 25.6?You got something like 25.6 m/s for the last part?
no mine was not this heheyeah
well but then the formula for power is differebt ib tht case its P=Fcos0 come on hows it possible to ignore he weight componentSorry to burst your bubble but the weight component shouldn't be taken. If you check may/june 2006 question 6 (iii), its pretty much the same math and only the resitive work done is taken.
Good and thanks god i didn't because im gving in m/j only this yearhow was it guys! i did well el7amdella was easy
weight component should be takenwell but then the formula for power is differebt ib tht case its P=Fcos0 come on hows it possible to ignore he weight component
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