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Sketch a pair of graphs to show the variation in kinetic and potential energies relative to height above x above the ground for an object dropped from 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 | 23 Sep, 2019, 08:20: PM
Kinetic energy K = (1/2) m v2 ................(1)

where m is mass of object, v is the velocity of object while falling down

velocity v of falling down object that is dropped from height is given by,

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

where x is the vertical distance above ground.

Using eqn.(1) and (2) , we get Kinetic energy K = (1/2)m (2g) (h-x) = m g (h-x) .................................(3)

Potential energy U at a distance x above ground,   U  = m g x  ...............................(4)

If potential energy and kinetic energy are equal, then from eqn.(3) and eqn.(4) we have ,   m g x = m g (h-x)  or   x = h/2

Hence when the falling object reaches half of the height , its potential energy is equal to kineteic energy

Answered by Thiyagarajan K | 24 Sep, 2019, 10:38: AM

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