Help please :)

- anonymous

Help please :)

- chestercat

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- anonymous

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- anonymous

Show that the gain in potential energy of the block is approximately 0.01J?

- anonymous

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## More answers

- 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

why is it 0.16

- anonymous

it is an approximation due to finding the height of the triangle

- anonymous

can u show the calculation?

- anonymous

how did u get mgh = 0.15?

- anonymous

as m=150/1000 kg
g=10 m/s^2
h=10/100 m
multiply them and get the ans

- anonymous

wt bt the other one 0.16

- anonymous

|dw:1350213752879:dw|
here h is slightly greater than the previous one thats why the ans will be 0.16 approx.

- anonymous

its 14.14 ? the height

- anonymous

how did u find the h

- anonymous

pythagoras....10^2 + 10^2 = x^2 x = 14.14

- anonymous

right, but how will u get the solu.

- anonymous

thats what i posted this question..

- anonymous

ok

- anonymous

how can the height be 11?

- anonymous

it was an appro

- anonymous

from 14?

- anonymous

was my method right

- anonymous

how did u get 11? i mean frm wt did u approximatre

- anonymous

it was just to get the ans dude

- anonymous

hmm...do we actually consider the height from the horizontal to cntre of gravity or as a whole?

- anonymous

can u give the solu

- 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

were u talking about GPE or only about PE

- anonymous

gain in potential energy?

- anonymous

i think the H should be taken from the cnt of gravity

- anonymous

so its 5 cm

- 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

yes :) so i got the answer as 0.03j not 0.01 can u chck where iwent wrong

- 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

so i am right :)

- Vincent-Lyon.Fr

You are!

- anonymous

i hve a second part in the question, cn i post it here..

- anonymous

go ahead

- 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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