1133(D)
Akash Anand Its not the right answer..try again.Upvote·0· Reply ·2013-03-13 22:37:52- Akash Anand Actually answer is given C but I am also getting D. You people try once again.
6 Answers
1133
99612k(5mg/k)2 -12k(3mg/k)2 =12mv2
from here we get v as 4g√mk
- Akash Anand You have not considered the gravitational potential energy.
229(D)
- Akash Anand Its not the right answer..try again.
- Akash Anand Actually answer is given C but I am also getting D. You people try once again.
1161Comment if I am wrong..
12K(6mgk)2=12K(2mgk)2+mg(8mgk)+12mv2
- Sourish Ghosh It should be mg*8mg/k.
- Swarna Kamal Dhyawala cant understand sir plzz explain it
- Dwijaraj Paul Chowdhury Still getting (D)
- Dwijaraj Paul Chowdhury Following sir's formula answer is zero o_O
- Swarna Kamal Dhyawala absolutely according to sir's eqn we get 0 and we correct it as mg*8mg/k then we get answer (D)
- Akash Anand @Sourish ..you are right...it was a typing mistake..now this is the correct expression.
- Akash Anand For those who did not get this expression..
In left hand side its the total initial spring potential energy (Since as equilibrium position spring was compressed already)
In right hand side this is the final spring potential energy(expansion in the spring can be found out from other condition in the question)+gain in gravitational potential energy(taking reference the compressed position as zero potential energy)+kinetic energy.
- Dwijaraj Paul Chowdhury But sir can we do this?
229let instantaneous extension be x' at time t=02
considering 3m block...having mg as normal reaction....
kx'+mg=3mg
kx'=2mg
x'=2mg/k
now...
(kx'^2)/2+(mv^2)/2 =0
taking mod value...nd solving v... we get v as 2g root (m/k)
answer--C
- Akash Anand Why you are equating whole thing with zero..that step I guess is not appropriate
- Gaurav Gardi I have done by one more method it is coming (D) from that method also.
- Dwijaraj Paul Chowdhury at a final point for a couple oscillation ...when the spring returns to its normal condition the system is momentarily at rest bfore it again starts oscillating due to its inertia of motion which implies that both K.E and P.E is zero....applying conservation of energy...then sum total of P.E nd K.E at this point of couple oscillation is zero
@Sir whats wrong with this?