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A person drops a $10 kg$ rock from the top of a $5 m$ ladder. The kinetic energy of the rock before it touches the ground is [[1]] $J$ with a speed of [[2]] $m / s$ .

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Detailed Solution

When a person drops a

10 kg

rock from the top of a

5 m

ladder, the kinetic energy of the rock before it touches the ground is

500 J

with a speed of

10 m / s

.
We are given that mass of the rock,

(m) = 10 kg

, and
height from which the rock is dropped,

(h) = 5 m

.
Using law of conservation of energy, we can say that

Loss of potential energy (P . E .) = gain in kinetic energy (K . E .)

Therefore,

K . E = mgh

= 10 \times 10 \times 5

\Rightarrow 500 J

Also, 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 velocity. Therefore,

\frac{1}{2} m v^{2} = 500 J

\Rightarrow \frac{1}{2} \times 10 \times v^{2} = 500

\Rightarrow v = 10 m / s

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