## zaphod 2 years ago Help please :)

1. zaphod

2. zaphod

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

3. zaphod

@Coolsector

4. zaphod

@sauravshakya @experimentX

5. rvgupta

|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

6. zaphod

why is it 0.16

7. rvgupta

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

8. zaphod

can u show the calculation?

9. zaphod

how did u get mgh = 0.15?

10. rvgupta

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

11. zaphod

wt bt the other one 0.16

12. rvgupta

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

13. zaphod

its 14.14 ? the height

14. rvgupta

how did u find the h

15. zaphod

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

16. rvgupta

right, but how will u get the solu.

17. zaphod

thats what i posted this question..

18. rvgupta

ok

19. zaphod

how can the height be 11?

20. rvgupta

it was an appro

21. zaphod

from 14?

22. rvgupta

was my method right

23. zaphod

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

24. rvgupta

it was just to get the ans dude

25. zaphod

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

26. rvgupta

can u give the solu

27. zaphod

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

28. rvgupta

29. zaphod

gain in potential energy?

30. rvgupta

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

31. zaphod

so its 5 cm

32. 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.

33. zaphod

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

34. 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

35. zaphod

so i am right :)

36. Vincent-Lyon.Fr

You are!

37. zaphod

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

38. rvgupta

39. zaphod

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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