how was M1 42

[mehdi1028]

it is 5N for the tension

No it was 4.8N

 

[mhalvi]

all answers are known now

 

[mhalvi]

HOW?\

WELL GUYS FOR THIS QUES WE WERE GIVEN ACCELERATIONS AT TOP AND BOTTOM RITE FOR EG LETS TAKE ONE SPEED THE ONE AT BOTTOM WE HAD TO CALCULATE ITFIRST
LOOK WE APPLIED F=MA RITE BY WHICH F-1000(RESISTIVE FORCE)=1250x4 F=6000 THAT IS THE DRIVING FORCE NOW FORMULA POWER=FORCEXVELOCITY 30000W/6000=5M/S THAT IS THE VELOCITY AT BOTTTOM THEN FOR THE SAME AT THE TOP BUT THIS TIME USE A AS 0.2 THEN SOLVE IT

I think you are missing one thing from your equation mgsinO becaause it was a hill and its force had to be taken into account in the equation.

 

[mhalvi]

I think you are missing one thing from your equation mgsinO becaause it was a hill and its force had to be taken into account in the equation.

and mgsinO was exactly 750N

 

[daviruss]

and mgsinO was exactly 750N

i love you man

 

[daviruss]

For B: T - 10(1-m) = a(1-m) <--------you forgot this
T - 10 +10m = 2 - 2m
T = 12 - 12m

Replace in A
8m = 12 - 12m
20m = 12
m = 0.6 kg

hmm thx for clearing this up. yup my fault srry guys but how many marks will i loose ? DAMN ! frgt this

 

[mhalvi]

i love you man

hehe thank you

 

[doorman1995]

how did you get the value of coefficient of friction in Q5? mine was .08 something
and work done in Q6?

well can u cnfrm me abt all my answers my Q6 is wrong wht abt the else ? hmm n for coefficient pasrt from rite we had angle 30 in part a hmm then we calculated the forward component of weight we had the acceleration so it was forward-friction=mxa then we got friction then divide by normal component if u have dont the same still answers diffeerent then post me q values ll do it

 

[doorman1995]

how did you get the value of coefficient of friction in Q5? mine was .08 something
and work done in Q6?

n my q 6 is wrong well i didnt add up the x component of weight so how many marks i wld loose? in later in part two ?

 

[mhalvi]

n my q 6 is wrong well i didnt add up the x component of weight so how many marks i wld loose? in later in part two ?

not many,dont worry.you just missed out one value

 

[Hassaan Khalid]

1. 1090J
2. a=2m/s2 m=0.6kg T=4.8N
3. 10m/s a=0.4m/s2
4. 39N direction 22.6' AC from x direction
5. 0=30' coefficient=0.693
6. gain in KE=344000J driving force workdone= 1220000J
7. distance OA=216m a=-0.72m/s2 u=17.6m/s

i think dat coefficient of friction was "0.9"

 

[doorman1995]

if u c

i think dat coefficient of friction was "0.9"

an get me all values i can tell the answer b calculations

 

[Hassaan Khalid]

if u c

an get me all values i can tell the answer b calculations

i dont remember all the values.....
u=4.4 m/s
v=0
s=65m
m was not given
t= (i dont remember, u?)
angle was 30

 

[Assad Ali]

i dont remember all the values.....
u=4.4 m/s
v=0
s=65m
m was not given
t= (i dont remember, u?)
angle was 3o

man how did u solve that kinectic energy question...did u consider the weight component in frictional force or not....cuz last time it was the same qustion and i forgot to take that in account and got it right...this time i was confused...?

 

[doorman1995]

well u r talking abt mayjune 2012 paper? i ignored the weight x component n took friction directly n i dont get it u mean that u had not calculated x component of weight separately and got ques rite?

 

[Hassaan Khalid]

i think that weight component is not taken into account while calculating energy....
during the paper i first calculated it with the weight component and the speed at bottom and at top were both 0.14 m/s
mass was only given to calculate energy

 

[mhalvi]

i think that weight component is not taken into account while calculating energy....
during the paper i first calculated it with the weight component and the speed at bottom and at top were both 0.14 m/s
mass was only given to calculate energy

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

 

[Assad Ali]

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

i dont think you are right either cuz the requirement never demanded us to consider that component but if in case we do consider it then what was the frictional force given for?

 

[Assad Ali]

i think that weight component is not taken into account while calculating energy....
during the paper i first calculated it with the weight component and the speed at bottom and at top were both 0.14 m/s
mass was only given to calculate energy

yep i did it this way but most people did mistake like they did the last time...

 

[Just visiting]

24 hrs passed right?......any way
if someone got a B in pure and an A in mechanics will he be able to get an overall A???
and what if i got 1 or 2 marks less than thresholds, will it be a B or an A?
screwed up in pure