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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 4Mg/k 2Mg/k Mg/k Mg/2k

Bhatoya Jatin , 7 Years ago

Grade 11

3 Answers

naman kumar singh

Last Activity: 7 Years ago

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

Aryan

Last Activity: 7 Years ago

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

Rishi Sharma

Last Activity: 4 Years ago

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