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