An object is dropped from h height. At what height will its kinetic energy be equal to its potential energy?
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By the law of conservation of energy, energy can neither be created nor destroyed. It can only transform from one form to another.
Since the motion of this object is happening only under gravity, the only energies it has are potential and kinetic, and the total energy is always conserved.
At height h, let PE be mgh. At that point, the body is not moving, so KE = 0.
At a height x, the PE will be mgx. Since at this point, KE is equal to PE, KE will also be mgx.
PE + KE = PE + KE
mgh + 0 = mgx + mgx
mgh = 2mgx
x = h/2
At h/2 height from the surface, the KE and PE of the object will be equal.
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Here's your answer...
By the law of conservation of energy, energy can neither be created nor destroyed. It can only transform from one form to another.
Since the motion of this object is happening only under gravity, the only energies it has are potential and kinetic, and the total energy is always conserved.
At height h, let PE be mgh. At that point, the body is not moving, so KE = 0.
At a height x, the PE will be mgx. Since at this point, KE is equal to PE, KE will also be mgx.
PE + KE = PE + KE
mgh + 0 = mgx + mgx
mgh = 2mgx
x = h/2
At h/2 height from the surface, the KE and PE of the object will be equal.
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