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SKETCH A PAIR OF GRAPHS TO SHOW THE VARIATION OF KINETIC AND POTENTIAL ENERGIES RELATIVE TO HEIGHT X ABOVE THE GROUND FOR AN OBJECT DROPPEDFROM REST FROM A HEIGHT h ABOVE THE GROUND. AT WHAT HEIGHT ABOVE THE GROUND WOULD THE TWO ENERGIES BE EQUAL TO EACH OTHER?
Asked by cbiswajit | 17 Sep, 2019, 08:36: PM
Kinetic energy KE = (1/2) mv2.........................(1)

where m is mass of the object and v is velocity of the object

when the object is dropped from a height h, velocity v as a function of height x above ground is given by

v2 = 2 g (h-x) ............................(2)

using eqn.(1) and (2), we write,  Kinetic energy KE = (1/2) m [ 2 g (h-x) ] = m g (h-x)   ......................(3)

Potential energy  PE = m g x ..........................(4)

Kinetic energy equation (3) and potential energy equation (4) are plotted in the figure.

if kinetic energy and potential energy are equal, then we have,  m g(h-x) = mg x  ....................(5)

FRom eqn.(5) we get ,  kinetic energy and potential enegy are equal when x = h/2
Answered by Thiyagarajan K | 18 Sep, 2019, 07:40: AM

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