A 12 kg block 0.4 m above an uncompressed spring is released from rest. What is the maximum
compression of the spring assuming it has a spring constant of 162 N/m?
I am trying to study for an exam and I need help with vertical springs. This problem was on a past exam and I am trying to figure out the best way to solve it.
So far I have
.5mv^2 + .5ky_e^2 + mgy_2^2 =.5mv_1^2 + .5k(y_1-y_e)^2 + mgy_1^2
but I do not believe that is correct
compression of the spring assuming it has a spring constant of 162 N/m?
I am trying to study for an exam and I need help with vertical springs. This problem was on a past exam and I am trying to figure out the best way to solve it.
So far I have
.5mv^2 + .5ky_e^2 + mgy_2^2 =.5mv_1^2 + .5k(y_1-y_e)^2 + mgy_1^2
but I do not believe that is correct
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1.48m is the answer
1)assume the bottom of spring to be the surface level
2)now the potential energy of the block is given by mgh (where h is the height,g the acc. due to earth=10m/s and m the mass of the object)
3)whole of this energy is stored in the spring at the time of maximum compression
4)applying conservation of energy
1/2k(x)^2 +mg(h-x) = mgh
solve for x
1)assume the bottom of spring to be the surface level
2)now the potential energy of the block is given by mgh (where h is the height,g the acc. due to earth=10m/s and m the mass of the object)
3)whole of this energy is stored in the spring at the time of maximum compression
4)applying conservation of energy
1/2k(x)^2 +mg(h-x) = mgh
solve for x
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i think the answer is 1.08 or -o.2
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the formula is; mg(x +y) =o.5 kx^2 (equation1) before that spring constant k=(mg/x) then, 0.5ky^2 -mgy-mgx=0.
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b