Mechanics 1 P42 2014; Discussion!

[Namehere]

Namehere

Hallelujah! We're not alone!!

Looks we´ve got some company! Though, not sure they did as bad as me! ^^

 

[thementor]

I think that you have a mistake in question 6.
The answer to part one should be 17.5, as net force will be equal to 15.5 so upthrust - weight = 15.5.
Weight would not stop acting when it enters liquid.

No the answer is 15.5 N.
net froce is equal to 13.5 N

 

[A star]

No the answer is 15.5 N.
net froce is equal to 13.5 N

No the answer is 15.5 N.
net froce is equal to 13.5 N

my ans was same but every one says its wrong :O if thats correct i mght scrape 46

 

[usama321]

No the answer is 15.5 N.
net froce is equal to 13.5 N

What about the weight then? Had they given us some data about the sides of the object, density of water etc, then we would have considered the resultant of weight - upthrust. However there was no such data. How would you propose then that we should just ignore the weight?

 

[usama321]

From what i know, you can write the equation in two ways
You can use
R - W = ma if you want to put deacceleration as positive value
W - R = -ma You put the value of a to be negative.
Both give 17.5N

 

[xtremeuser12345]

What about the weight then? Had they given us some data about the sides of the object, density of water etc, then we would have considered the resultant of weight - upthrust. However there was no such data. How would you propose then that we should just ignore the weight?

u already know that from R since R is the net frictional and viscous forces on the object calced before this q to be 15.5 N.after this to find tension:T-mg-R=ma which gives T=17.59 N.

 

[usama321]

u already know that from R since R is the net frictional and viscous forces on the object calced before this q to be 15.5 N.after this to find tension:T-mg-R=ma which gives T=17.59 N.

Tension? What does tension have to do with our question? Furthermore, there is difference between upward thrust and viscous forces. They just told us to find the frictional forces from what i remember

 

[Oishee Asif]

Can we stop the discussion now? ;_;
Let's just wait until the office a MS comes out?

 

[usama321]

Can we stop the discussion now? ;_;
Let's just wait until the office a MS comes out?

I think i'll just shut xpc down for a few days :3 Otherwise, all these different answers are gonna make me go mad xD

 

[raysonzaffar]

Can we stop the discussion now? ;_;
Let's just wait until the office a MS comes out?

I thank this thread for increasing my depression levels. i was expecting a 45 after i came out of the exam hall. Now I'll be seen in the November session

 

[Oishee Asif]

Loool.
:\

 

[xtremeuser12345]

I think i'll just shut xpc down for a few days :3 Otherwise, all these different answers are gonna make me go mad xD

do u know when a MS is going to be uploaded?

 

[Oishee Asif]

do u know when a MS is going to be uploaded?

Yeah, July.

 

[xtremeuser12345]

Yeah, July.

then 2 months from now?where can i find them once they are out?

 

[thementor]

From what i know, you can write the equation in two ways
You can use
R - W = ma if you want to put deacceleration as positive value
W - R = -ma You put the value of a to be negative.
Both give 17.5N

So W = 2N and ma = (0.2)(67.5) = 13.5
so R = W + ma = 2 + 13.5 = 15.5
i am sorry if I sound sarcastic, but where is the 17.5?
I sued the equation u suggested and I stil got 15.5

We DID take the weight into account.

 

[Oishee Asif]

then 2 months from now?where can i find them once they are out?

Depends on the dude who uploads them. They'll be on olevel.sourceforge.net when he updates it.

 

[usama321]

So W = 2N and ma = (0.2)(67.5) = 13.5
so R = W + ma = 2 + 13.5 = 15.5
i am sorry if I sound sarcastic, but where is the 17.5?
I sued the equation u suggested and I stil got 15.5

We DID take the weight into account.

K seriously! Facepalm :3 I just checked my answer list and i have 15.5 I think I have been mixing it up with the second part Sorry Sorry Sorry

 

[thementor]

K seriously! Facepalm :3 I just checked my answer list and i have 15.5 I think I have been mixing it up with the second part Sorry Sorry Sorry

hahahah no worries. man Chill. Just relax. Calm down. Enjoy and good luck for ur next papers.

 

[Oishee Asif]

Thank you very much.

 

[shakky]

-6

are you sure it was -6? i just asked the ques from my sir and he said that when B reaches the floor the o.25 kg mass travels up due opto the velocity gained by it hence the tensions in both the strings become slack.. therefore the deceleration of the 0.5 kg mass (p) would be equal to Reaction force into coefficient of friction i.e 5 x 0.4 = (0.5) a. which gives a to be 4 .. ps my answer was -6 too . and do you by any chance remember what the ques was worth becuase i dont think it was worth only two marks..