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A body of mass $2 kg$ falls from the rest. The kinetic energy during the fall at the end of $2 s$ is [[1]] $J$ .

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Detailed Solution

When a body of mass

2 kg

falls from the rest then the kinetic energy during the fall at the end of

2 s

400 J

.
The formula for kinetic energy

(K . E .)

can be expressed as

K . E . = \frac{1}{2} m v^{2}

where

m

is the mass of the object and

v

is its final velocity.
We are given that,
Mass of the body

(m) = 2 kg

Time taken

(t) = 2 s

Since the object is released from rest,
Initial velocity

(u) = 0 m / s

gravitational acceleration

(g) = 10 m / s^{2}

Velocity at the end of

2 s

, is calculated using the first equation of motion,

v = u + gt

\Rightarrow v = 0 + 10 \times 2

\Rightarrow v = 19.6 m / s

v = 20 m / s

So the kinetic energy at the end of

2 s

K . E . = \frac{1}{2} \times 2 \times 20^{2}

\Rightarrow K . E . = 400 J

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