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Explainthe conservative of mechanical energy for a free fall body . |
Answer» Solution :Let a ball of mass m being dropped from a cliff of height H as shown in figure .  The total mechanical energies `E_(0),E_(h) and E_(H)` of the ball at the indicated heights , zero (ground level ) h and H respectively are . Total mechanical ENERGY at height H , `E_(H) =mgH +1/2 mv^(2)` but a maximum height v = 0 Total energy at h height , `E_(h)` = POTENTIAL energy + kinetic energy ` = mgh+1/2 mv_(h)^(2) ( :."velocity of ball at height h = "v_(h))` Total energy at ground , `E_(0) =/2 mv_(f)^(2)` where `v_(f)` is the final velocity of ball when it hit the ground . From the conservation of mechanical energy at height H , `E_(H) =E_(0)` ` :.mgH = 1/2 mv_(f)^(2)` where `v_(f)` is the final velocity of ball when it hit the ground . From the conservation of mechanical energy at height H , `E_(H) =E_(0)` ` :. mgH = 1/2 mv_(f)^(2)` ` :. v_(f) = sqrt(2gH)` From the conservation of mechanical energy at H and h . `E_(H) =E_(h)` ` :. mgH =mgh + 1/2 mv_(f)^(2)` ` :. mgH =mgh +1/2mv_(f)^(2)` ` :. GH -gh =1/2 v_(f)^(2)` ` :. 2g(H-h)=v_(f)^(2)` ` :. v_(f)=sqrt(2g(H-h))` At the height H , the energy is purely potential energy at height h and is fully kinetic at ground level . HENCE , this illustrates the conservation of mechanical energy for a free fall body . |
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