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

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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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Gravitational potential converts to spring potential as block slides and spring compresses. So balance the 2 taking account of slope.
Potential energy of compressed spiring=.5kx^2=.5*10^4*.04^2=?
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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Other answers:

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

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