All Mechanics DOUBTS HERE!!!

[hesho21]

yeh sorry missed out the (ii) part, but anyways u got it Good luck

[solo_G]

right okay,
so in part (ii) you found the height above the ground of P and Q, and their speed, when the string breaks.
that's s_P= 3 m, s_Q= 7 m, and their speed is 2 ms^-1.
you use the equation s = ut + 1/2 at², except here they're accelerating due to gravity only so a=10
so for P:
3 = 2t + 1/2 (10)t²
solving for t you get t = 0.6 or t = -1 (obviously the second is invalid)
for Q:
7 = -2t + 1/2 (10)t²
here it's -2 because it's in the opposite direction to P. remember this is velocity, you have to take into consideration the speed AND the direction.
so yeah, solve for this one you get t=1.4 or t=-1

1.4 - 0.6 = 0.8

yeah i know about the solving but the thing that i dont get is that when the string breaks both particles will move downwards and so whats the point of using u=-2.....soory dude bt am lil bit bad at mechanics

[falafail]

yeah i know about the solving but the thing that i dont get is that when the string breaks both particles will move downwards and so whats the point of using u=-2.....soory dude bt am lil bit bad at mechanics

oh right sh*t my bad. it's not negative because they're in opposite directions, they're in the same direction -_- sorry.
okay so you know that velocity is a vector quantity; you take into consideration the speed AND the direction.
here the direction matters especially because, in the first parts of the question, we took the downwards velocity of P as positive and UPWARDS of Q as positive.
so in this part, since P is going down and so is Q, the velocity of P remains positive as it's still int he downwards direction, but the velocity of Q becomes negative, since it is now going downwards (opposite to what we took as positive)
geddit?

[Saladin]

oh right sh*t my bad. it's not negative because they're in opposite directions, they're in the same direction -_- sorry.
okay so you know that velocity is a vector quantity; you take into consideration the speed AND the direction.
here the direction matters especially because, in the first parts of the question, we took the downwards velocity of P as positive and UPWARDS of Q as positive.
so in this part, since P is going down and so is Q, the velocity of P remains positive as it's still int he downwards direction, but the velocity of Q becomes negative, since it is now going downwards (opposite to what we took as positive)
geddit?

You did not have to use that word did you?

[falafail]

You did not have to use that word did you?

what word? >__>

[solo_G]

oh right sh*t my bad. it's not negative because they're in opposite directions, they're in the same direction -_- sorry.
okay so you know that velocity is a vector quantity; you take into consideration the speed AND the direction.
here the direction matters especially because, in the first parts of the question, we took the downwards velocity of P as positive and UPWARDS of Q as positive.
so in this part, since P is going down and so is Q, the velocity of P remains positive as it's still int he downwards direction, but the velocity of Q becomes negative, since it is now going downwards (opposite to what we took as positive)
geddit?

Aye aye i got that!!! now it makes sense....thanks alot falafail  for ur help!

[falafail]

Aye aye i got that!!! now it makes sense....thanks alot falafail  for ur help!

no problem

[AL*Eagle]

I have the answer of this question.. but i dont get it..
would someone care to explain?

A 5Kg box lies on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.5 .
A force P is applied to to the box to pull or push it horizontally along the floor.
Find the Magnitude of P which is necessary to achieve this if:
a P is applied at an angle x above the horizontal, where tan x = 3/4
b P is applied at an angle x below the horizontal, where tan x = 3/4

Id really appreciate anyone explainin this??

[falafail]

I have the answer of this question.. but i dont get it..
would someone care to explain?

A 5Kg box lies on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.5 .
A force P is applied to to the box to pull or push it horizontally along the floor.
Find the Magnitude of P which is necessary to achieve this if:
a P is applied at an angle x above the horizontal, where tan x = 3/4
b P is applied at an angle x below the horizontal, where tan x = 3/4

Id really appreciate anyone explainin this??

is it P > 19.2 N for part a and P > 35.7 N for part b? i'll explain how i got those if i'm right

[astarmathsandphysics]

try this

[immortal]

Wts da ans, because i get P>22.7 for a   &   P>125  for b

[AL*Eagle]

is it P > 19.2 N for part a and P > 35.7 N for part b? i'll explain how i got those if i'm right

Answers are
a 22N
b 49
(2s.f.)
close to yours

[AL*Eagle]

try this

thanks for your effort man

[falafail]

Answers are
a 22N
b 49
(2s.f.)
close to yours

right well i found out what i did wrong, but it's still not what you have. maybe you rounded wrong? 'cause i get 22.7 N and 50 N (exact answers are somewhere around 22.727.... and 50.000... so i'm pretty sure i'm right )

[immortal]

right well i found out what i did wrong, but it's still not what you have. maybe you rounded wrong? 'cause i get 22.7 N and 50 N (exact answers are somewhere around 22.727.... and 50.000... so i'm pretty sure i'm right )

I did da same, wat u did wrong waz 2 tke g as 10 & not 9.8