CBSE Class 11-science Answered
show that the total mechanical energy of a body falling under gravity in conserved discuss the graphically also
Asked by satya785583 | 22 Mar, 2019, 07:26: AM
Expert Answer
Let us assume an object of mass m is dropped from a height h above ground level.
initially just before dropping this object does not have any kinetic energy, it has only potential energy.
Hence total energy = kinetic energy + potential energy = m×g×h ...............................(1)
where g is acceleration due to gravity.
Let us calculate the total energy at a height y above ground level, in between the initial height h and ground level.
potential energy = m×g×y ...........................(2)
kinetic energy = (1/2)×m×v2 ........................(3)
where v is the speed attained by the object at the height y.
Since the distance travelled by the object is (h-y) we get from equation of motion, v2 = 2×g×(h-y)
by substituting for v2 in eqn.(3) we get kinetic energy = (1/2)×m×2×g×(h-y) = m×g×(h-y) ..................(4)
By adding eqn.(3) and eqn.(4), we get total energy = m×g×y + m×g×(h-y) = m×g×h ...........(5)
When the object reaches ground level, its potential energy is zero .
At ground level, just before hitting the ground, kinetic energy = (1/2)×m×v2 = (1/2)×m×(2×g×h) = m×g×h ......................(6)
From eqn.(1), (5) and (6), we can conclude that total energy m×g×h always remains same, hence total energy is conserved.
Answered by Thiyagarajan K | 22 Mar, 2019, 09:23: AM
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