Comments on: Problem? with energy and work? http://www.renewable-energy-at-home.com/uncategorized/problem-with-energy-and-work How you can save with renewable energy at home Thu, 10 Jun 2010 07:41:20 +0000 hourly 1 http://wordpress.org/?v=3.0 By: Ahcai http://www.renewable-energy-at-home.com/uncategorized/problem-with-energy-and-work/comment-page-1#comment-760 Ahcai Sun, 24 May 2009 21:34:59 +0000 http://www.renewable-energy-at-home.com/7/problem-with-energy-and-work#comment-760 a) Lets assume that the vertical height of the bob is H and let the vertical distance from the bob to the point of attachment of the string to the ceiling be X. X/L = cos 25 X = L cos25 H = L - Lcos25 = L(1-cos25) (SHOWN) b) The height when the bob is at an angle of 9 degrees, H' = L (1-cos9) = 0.75(1 - cos 9 ) The kinetic energy = mgH - mgH' = 0.15(9.81)[0.75(1 - cos25)] - 0.15(9.81)[0.75(1 - cos9)] = 0.0898 J c) By conservation of energy, mgH = 1/2mv^2 gH = 1/2(v^2) v^2 = 2gH = 1.379 v = 1.17m/s Hope this is correct ^^<br><b>References : </b><br> a) Lets assume that the vertical height of the bob is H and let the vertical distance from the bob to the point of attachment of the string to the ceiling be X.

X/L = cos 25
X = L cos25
H = L – Lcos25 = L(1-cos25) (SHOWN)

b) The height when the bob is at an angle of 9 degrees, H' = L (1-cos9) = 0.75(1 – cos 9 )

The kinetic energy = mgH – mgH' = 0.15(9.81)[0.75(1 - cos25)] – 0.15(9.81)[0.75(1 - cos9)] = 0.0898 J

c) By conservation of energy,

mgH = 1/2mv^2
gH = 1/2(v^2)
v^2 = 2gH = 1.379
v = 1.17m/s

Hope this is correct ^^
References :

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