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A mass $‘ m ’$ falls from a height $‘ h ’$ . At any point on its path the total energy is

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\frac{mgh}{2}

\frac{mgh}{4}

mgh

Depends upon the height

answer is C.

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Detailed Solution

A mass

‘ m ’

falls from a height

‘ h ’

. At any point on its path, the total energy is

mgh

.
According to the law of conservation of energy the total energy of the mass will be equal to its potential energy at the maximum height, i.e.,

mgh

.
where,

m

is the mass of the object,

g

is the acceleration due to gravity, and

h

is the height of the object.
Question Image

P.E at ‘A’

= mgh

K E at ‘A’

= 0

∴ Total energy at A

= mgh

As it starts falling at the point ‘B’ it would have lost some

P . E

and gained some

K . E

. we know that

v^{2} = u^{2} + 2 as

where

v =

final velocity

u =

initial velocity

a =

acceleration (

g = 9.81 m / s^{2}

in this case)

s =

distance travelled (assumed

x

in this case)
∴

{V^{2}}_{B} = 0 + 2 g x

(as initial velocity

u = 0

)
∴

V_{B} = \sqrt (2 gx)

P . E

at B

= mg (h – x)

K . E

at B

= \frac{1}{2} m {(\sqrt{2 gx})}^{2}

\Rightarrow

= \frac{1}{2} m \times 2 \times g \times x

\Rightarrow

= mgx

∴ Total energy at B

= mgh – mgx + mgx

Total energy = mgh

Thus total energy at any point in its path is

mgh

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