Question

A stone is dropped from a height h. Its potential energy at
from a height h. Its potential energy at a particular instant is
nine times its kinetic energy.
When the velocity of the stone becomes double the velocity at that instant
will be the ratio of its new kinetic energy to new potential energy?​

Answer 1

Answer:

2:3

Explanation:

Let the distance through which the stone dropped = d

Now,

it is given that at 'd', potential energy = 9 × Kinetic energy

Potential energy when the stone dropped off to 'd' = mg(h-d)

where, m is the mass of the stone

g is the acceleration due to the gravity

also,

Now gain in kinetic energy while dropped through distance 'd' = loss in potential energy

or

Kinetic energy = mgd

now using the relation given in the problem

mg(h-d) = 9 × mgd

or

h - d = 9d

or

d = h/10

therefore,

the kinetic energy for the first case = mgd = mg(h/10)

also

Kinetic energy = $\frac{1}{2}mv^2$ = mg(h/10)

where, v is the velocity

and potential energy = 9 × Kinetic energy = 9 × mg(h/10) = (9/10)mgh  .....(1)

now,

for the case 2

velocity of the stone = 2 times the velocity in case 1

or

v' = 2v

thus,

the kinetic energy = $\frac{1}{2}mv'^2$ = $\frac{1}{2}m(2v)^2$

or

the kinetic energy = $4\frac{1}{2}mv'^2$

or

the kinetic energy = 4×mg(h/10)

thus,

the potential energy at this point = Initial potential energy - Kinetic energy

or

the potential energy at this point = mgh - 4×mg(h/10) = (6/10)mgh  .......(2)

dividing 2 by 1, we have

PE₂/PE₁ = $\frac{ (6/10)mgh}{(9/10)mgh}$

or

Ratio of new potential energy to old = 2:3