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An ideal spring with spring constant k is hung from the ceiling and a block of mass M is hung to its its lower end. The mass is released with spring initially unstretched. Then the maximum extending in the spring is
  1. 4Mg/k
  2. 2Mg/k
  3. Mg/k
  4. Mg/2k

Bhatoya Jatin , 8 Years ago
Grade 11
anser 3 Answers
naman kumar singh
Answer will be 2mg/k firsrt spring stretches to its equi position given as Kx=mg implies x=mg/k now total energy at this point is total energy of shm that block performs.this point is equi point so energy here is 1/2kA*A which is equal to total energy and total energy is -mgx(work done by gravity) +mgx(=1/2mv*v as v=underoot 2gx)+1/2kx*x=1/2kA*A which gives A=mg/k so total elongation is A+x=2mg/k
Last Activity: 8 Years ago
Aryan
Ans: work done by gravity plus work done by spring = change in kinetic energy which is zero From Work-energy theorem; Wg+Ws=0⇒mgx−(1/2)kx2=0⇒x=2mg/k
Last Activity: 8 Years ago
Rishi Sharma
Dear Student,
Please find below the solution to your problem.

Firsrt spring stretches to its equi position given as
Kx=mg
implies
x=mg/k
now total energy at this point is total energy of shm that block performs.
this point is equi point so energy here is 1/2kA*A
which is equal to total energy and total energy is
-mgx(work done by gravity) +mgx(=1/2mv*v as v=underoot 2gx)+1/2kx*x
=1/2kA*A
which gives A=mg/k
so total elongation is A+x=2mg/k

Thanks and Regards
Last Activity: 5 Years ago
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