anonymous
  • anonymous
Help please :)
Physics
chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
Show that the gain in potential energy of the block is approximately 0.01J?
anonymous
  • anonymous

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anonymous
  • anonymous
anonymous
  • anonymous
|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
anonymous
  • anonymous
why is it 0.16
anonymous
  • anonymous
it is an approximation due to finding the height of the triangle
anonymous
  • anonymous
can u show the calculation?
anonymous
  • anonymous
how did u get mgh = 0.15?
anonymous
  • anonymous
as m=150/1000 kg g=10 m/s^2 h=10/100 m multiply them and get the ans
anonymous
  • anonymous
wt bt the other one 0.16
anonymous
  • anonymous
|dw:1350213752879:dw| here h is slightly greater than the previous one thats why the ans will be 0.16 approx.
anonymous
  • anonymous
its 14.14 ? the height
anonymous
  • anonymous
how did u find the h
anonymous
  • anonymous
pythagoras....10^2 + 10^2 = x^2 x = 14.14
anonymous
  • anonymous
right, but how will u get the solu.
anonymous
  • anonymous
thats what i posted this question..
anonymous
  • anonymous
ok
anonymous
  • anonymous
how can the height be 11?
anonymous
  • anonymous
it was an appro
anonymous
  • anonymous
from 14?
anonymous
  • anonymous
was my method right
anonymous
  • anonymous
how did u get 11? i mean frm wt did u approximatre
anonymous
  • anonymous
it was just to get the ans dude
anonymous
  • anonymous
hmm...do we actually consider the height from the horizontal to cntre of gravity or as a whole?
anonymous
  • anonymous
can u give the solu
anonymous
  • anonymous
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
anonymous
  • anonymous
were u talking about GPE or only about PE
anonymous
  • anonymous
gain in potential energy?
anonymous
  • anonymous
i think the H should be taken from the cnt of gravity
anonymous
  • anonymous
so its 5 cm
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Gain in potential energy is m g Δh , where Δh is gain in altitude of the centre of mass of the body.
anonymous
  • anonymous
yes :) so i got the answer as 0.03j not 0.01 can u chck where iwent wrong
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Δh = 5√2 - 5 = 2.07 cm = 0.0207 m g = 9.81 m/s² m = 0.150 kg Hence gain = 0.03 J
anonymous
  • anonymous
so i am right :)
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
You are!
anonymous
  • anonymous
i hve a second part in the question, cn i post it here..
anonymous
  • anonymous
go ahead
anonymous
  • anonymous
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
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