Given:
The height from which an object is dropped is: h = 6 m
Potential energy (PE) = Kinetic energy (KE)
To find:
The height at which the potential energy is equal to kinetic energy.
Step-by-step explanation:
Using the law of conservation of the energy, we get:
TE(top) = TE(at height h')
The total energy at the top will be the addition of potential energy "mgh" and kinetic energy. But at the top, the kinetic energy of an object will be zero. Thus, we get:
PE(top) + KE(top) = PE + KE
mgh + 0 = PE + KE
But, it is given that, at height h', PE = KE. Thus, we get:
mgh = PE + PE
mgh = 2PE
mgh = 2mgh'
Here, h' is the height at which PE = KE
h = 2h'
Rearranging the above equation, we get:
h' = h/2
Substituting the values in the above equation, we get:
h' = 6m/2 = 3 m.
Final answer:
Thus, at the height of 3 m, the kinetic energy of an object will be equal to its potential energy.