1
prateek punj
·Mar 25 '09 at 8:05
hey buddy at a particular instant of time the net force acting on each box will be different at each moment then each block will acelerate at every instant of time...there acceleration will keep on accelerating....
33
Abhishek Priyam
·Apr 2 '09 at 4:24
v 2kg=5t2/2-10t+10 for t>2 sec
v 3kg=5t2/3-10t+15 for t>3 sec
v2kg at t=5 =45/2
v3kg at t=5 =20/3
vcom=2*(45/2)+3*(20/3) = 13m/s
2+3
0∫sds=2∫55t2/2-10t+10 dt=135/6 ..(for 2 kg)
0∫sds=3∫55t2/3-10t+15 dt=40/9 ..(for 3kg)
ycom=(135/6)*2+(40/9)*3
2+3
=35/3=11.667
3
iitimcomin
·Apr 2 '09 at 2:09
between 2 and 3 s the 2kg block will move up .......
10t - 20 = 2dv/dt ........
5t^2 - 20t = 2v .......
5/3t^3 - 10t^2=2x
put imits frm 2 ->3 .....
thus find x .....
after that 20t-5d = 5 dv/dt....
integrate twice and find new height of com /////////////////////
3
iitimcomin
·Apr 2 '09 at 1:54
vel. of 2 kg block wen 3 kg block just takes off is .........
10t - 2g = 2 dv/dt ......
integratin frm 2 to 3 we get 5/2 m/s ..........................
now velocity of com at dat point iz .........2*5/2 + 0 /5 =1m/s..........................
now after dat entire sysm. moves ......
20t - 5g = 5 dv/dt v(com)!!!!!!!
integratin time from 3 to 5 and velocity from 1 to v ......
we het v as 13m/s!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
33
Abhishek Priyam
·Apr 1 '09 at 8:42
aa jaega waisa karne se... lekin man nahi kar raha banane ka.. :P actually aaj kal chem padh rahe hain [3]
33
Abhishek Priyam
·Apr 1 '09 at 8:37
What you have to do is..
Find velocity of lighter block when just heavier leave the floor.
Then Vcom=m1v1/(m1+m2)
there after when each block has left the floor.. forces on com are only gravity and 20t
so (m1+m2)dv/dt=20t-(m1+m2)g
limit from Vcom to v and time from t(at which last block had left the floor) to 5 sec.
this will give V com at t=5sec
Similarly for y coordinate.
1
vector
·Apr 1 '09 at 8:26
mujhe to bas ye pata chala hai ki lighter block ll leave d surface at2 seconds n heavier ll leave it at 2.4 seconds
1
vector
·Apr 1 '09 at 8:26
mujhe to bas ye pata chala hai ki lighter block ll leave d surface at2 seconds n heavier ll leave it at 2.4 seconds
1
Samarth Kashyap
·Apr 1 '09 at 6:52
since the pullety is massless 2T=20t
T=10t
the blocks will start acc. iff T≥2g ie 10t≥2g or t≥2s
so before t=2s vcm=0
after t=2s the 2kg block loses contact with the ground and the 3 kg block is still on the ground till t=3s(calculated as above)
so for 2≤t≤3 vcm=v 2kg
still not able to make out after t≥3s
1
prateek punj
·Apr 1 '09 at 6:00
yaar koi toh try karoo....[35]
1
prateek punj
·Apr 1 '09 at 5:37
someone watch this one also....
11
virang1 Jhaveri
·Mar 25 '09 at 8:03
The total pulley system is forced up wards therefore blocks does not matter
11
virang1 Jhaveri
·Mar 25 '09 at 8:02
S = ut + at2/2
S = 5.1*2.55*2.55/2
S= 16.5m
Sorry but the answer is not matching
1
prateek punj
·Mar 25 '09 at 8:01
hey buddy but the blocks will keep on aacelerating but u hav taken it to be constant....5.1 (calculated)
11
virang1 Jhaveri
·Mar 25 '09 at 7:58
F = 20t
Downward force = 5*9.8 = 49
therfore 20t = 2.45
t = 2.45 secs
Average F from 2.45 to 5 = 20*2.45 = 149/2 = 74.5
Total net fore up - down
= 74.5-49 = 25.5
Mass = 5kg
F = ma
a = 5.1
v = at
v = 5.1(2.55)
v = 13m/s
1
prateek punj
·Mar 25 '09 at 7:58
guys keep on trying i'll taake a look tomorrow....
1
prateek punj
·Mar 25 '09 at 7:47
anyone else please try it out yaar....
1
prateek punj
·Mar 25 '09 at 7:45
kuch bhi nahin hua kya.....
11
Subash
·Mar 25 '09 at 7:44
:(
no not getting
some one else will explain
1
prateek punj
·Mar 25 '09 at 7:43
subhash hav u done it....
1
greatvishal swami
·Mar 25 '09 at 7:16
[9][9]
k i m not trying now doin some imp work but bookmarked will try later
k
1
prateek punj
·Mar 25 '09 at 7:14
if u r trying it out just tell me if not then too.....
1
prateek punj
·Mar 25 '09 at 7:08
NONE TO ANSWER IT YAAR....