1hey 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....
In the figure shown surface of massless pulley is smooth. The string is massless. The two blocks are placed on a smooth horizontal surface . A time varying force F=20t is applied. Considering the two blocks and the pulley and the string as single system find ata t=5..
a) Velocity of the centre of mass of the system
b) y-coordinate of centre of mass of the system
PLEASE REPLY SOON......
ANS :- (a) 13 m/s (b)11.67 m
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UP 0 DOWN 0 5 30
30 Answers
1
33v 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
3between 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 /////////////////////
3vel. 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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
33aa jaega waisa karne se... lekin man nahi kar raha banane ka.. :P actually aaj kal chem padh rahe hain [3]
33What 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.
1mujhe to bas ye pata chala hai ki lighter block ll leave d surface at2 seconds n heavier ll leave it at 2.4 seconds
1mujhe to bas ye pata chala hai ki lighter block ll leave d surface at2 seconds n heavier ll leave it at 2.4 seconds
1since 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
11The total pulley system is forced up wards therefore blocks does not matter
11S = ut + at2/2
S = 5.1*2.55*2.55/2
S= 16.5m
Sorry but the answer is not matching
1hey buddy but the blocks will keep on aacelerating but u hav taken it to be constant....5.1 (calculated)
11F = 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[9][9]
k i m not trying now doin some imp work but bookmarked will try later
k