DominantVampire
  • DominantVampire
a block of mass 12 kg is released from rest om a friction-less incline of the angle theta=30 below the block is a spring having spring constant 10^4 N/m. the block stops momentarily when it compresses the spring by 4 cm. how far does the block move down the incline from its rest position to this stopping point
Physics
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SOLVED
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jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
Gravitational potential converts to spring potential as block slides and spring compresses. So balance the 2 taking account of slope.
shamim
  • shamim
Potential energy of compressed spiring=.5kx^2=.5*10^4*.04^2=?
shamim
  • shamim
Potential energy at the top of inclined plane=kinetic energy at the bottom of inclined plane=mgh=12*9.8*h

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shamim
  • shamim
Potential energy of spring=kinetic energy of block
shamim
  • shamim
h=?
shamim
  • shamim
At last u hv to find out the length of inclined plane
Pawanyadav
  • Pawanyadav
MgHsin30 ⏰=0.5×10^4×4^2
DominantVampire
  • DominantVampire
@sikinder is this formula above written right which as been suggested
DominantVampire
  • DominantVampire
but the ans is totally diff. the h i get is 0.136m
anonymous
  • anonymous
i was thinking that ur calculation method is right @DominantVampire
DominantVampire
  • DominantVampire
is it?
anonymous
  • anonymous
yes i think @DominantVampire
DominantVampire
  • DominantVampire
alright. thankyou.

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