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Question

If girl of mass $3.5 k g$ climbs up from the first floor of the building at a height $4 m$ above the ground to the third floor at height $12 m$ above the ground. What will be the increase in gravitational potential energy? (Take, $g = 10 m s^{2}$ )

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Solution

Given data:
Mass, $m = 3.5 k g$ ,
Gravitational force, $g = 10 m s^{2}$
Height at first floor, $h_{1} = 4 m$
Height at second floor, $h_{2} = 12 m$
Gravitational potential energy at $1^{s t}$ floor
$P E_{1} = m g h_{1}$
$P E_{1} = 3.5 \times 10 \times 4$
$P E_{1} = 140 J$
Gravitational potential energy at $2^{n d}$ floor
$P E_{2} = m g h_{1}$
$P E_{2} = 3.5 \times 10 \times 12$
$P E_{2} = 420 J$
Increase in potential energy
By differentiating the potential energy at $1^{s t}$ floor from potential energy at $2^{n d}$ floor we get the increase in potential energy.
$P E_{i n c r e a s e} = P E_{2} - P E_{1}$
$P E_{i n c r e a s e} = 420 - 140$
$P E_{i n c r e a s e} = 280 J$
The increase in gravitational potential energy is $280 J$

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