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zaphod
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zaphod
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Show that the gain in potential energy of the block is approximately 0.01J?
zaphod
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@Coolsector
zaphod
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@sauravshakya @experimentX
rvgupta
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|dw:1350213222437:dw|
now subtract them and get the answer
this difference is due to the increase in hight of second square due to being on its edge
zaphod
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why is it 0.16
rvgupta
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it is an approximation due to finding the height of the triangle
zaphod
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can u show the calculation?
zaphod
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how did u get mgh = 0.15?
rvgupta
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as m=150/1000 kg
g=10 m/s^2
h=10/100 m
multiply them and get the ans
zaphod
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wt bt the other one 0.16
rvgupta
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|dw:1350213752879:dw|
here h is slightly greater than the previous one thats why the ans will be 0.16 approx.
zaphod
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its 14.14 ? the height
rvgupta
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how did u find the h
zaphod
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pythagoras....10^2 + 10^2 = x^2 x = 14.14
rvgupta
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right, but how will u get the solu.
zaphod
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thats what i posted this question..
rvgupta
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ok
zaphod
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how can the height be 11?
rvgupta
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it was an appro
zaphod
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from 14?
rvgupta
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was my method right
zaphod
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how did u get 11? i mean frm wt did u approximatre
rvgupta
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it was just to get the ans dude
zaphod
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hmm...do we actually consider the height from the horizontal to cntre of gravity or as a whole?
rvgupta
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can u give the solu
zaphod
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i dont knw how to solve it, coz the answer doesnt equal to 0.01 j, maybe theres a problem in the question. anyway thanks for helping
rvgupta
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were u talking about GPE or only about PE
zaphod
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gain in potential energy?
rvgupta
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i think the H should be taken from the cnt of gravity
zaphod
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so its 5 cm
Vincent-Lyon.Fr
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Gain in potential energy is m g Δh , where Δh is gain in altitude of the centre of mass of the body.
zaphod
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yes :) so i got the answer as 0.03j not 0.01 can u chck where iwent wrong
Vincent-Lyon.Fr
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Δh = 5√2 - 5 = 2.07 cm = 0.0207 m
g = 9.81 m/s²
m = 0.150 kg
Hence gain = 0.03 J
zaphod
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so i am right :)
Vincent-Lyon.Fr
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You are!
zaphod
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i hve a second part in the question, cn i post it here..
rvgupta
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go ahead
zaphod
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The collision is perfectly elastic and, without sliding, the block turns about the corner as shown. The block is able to reach the position as shown in the image.
25% of the kinetic energy of the ball is transferred to the block.
Calculate
1. the kinetic enrgy of the balll just before it strikes the block