kindly help please. A Block Of Mass 12 kg

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kindly help please. A Block Of Mass 12 kg

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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
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no body knows perhaps, sighs.

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Other answers:

do you know how to use energy conservation??
yes.
Ki+Ui=Kf+Uf ?
yep... initial energy will be the potential energy of the block and the final energy will be the spring extension.. can you relate?
compression**
the Ki is 0 and Uf would be zero as well i guess
Uf would not be zero as the spring is compressed and is having potential energy its like the GPE is getting converted to spring energy
GPE is gravitational potential energy
yes.
the spring constant is 10,000N/m ?
yes
will it be mgh=1/2 kx^2
yes
and plug in the values? h is not given
h is what you have to find!!
v have to find out h?
yes, sorry
so it becomes like h = [(1/2)kx^2][mg]
correct?
is mg in the denom?
h = [(1/2)kx^2] / [mg]
then correct
the h is 6.80 m
now what
sin(30) = h/x to determine the hypotenuse
now h/sin30=x ?????
the ans is 13.6 ????

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