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

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

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..

ok

how can the height be 11?

it was an appro

from 14?

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

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 :)

You are!

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

go ahead

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