Question

If 5KJ of work is done in lifting 20kg of water to the top of a building , calculate the height of the building ( take g = 20 m /s)​

Answer 1

Solution :-

Work done = 5kJ

We know that, 1kJ = 1000J

Therefore, 5kJ = 5000J

Mass of water = 20kg

Value of g = 10m/s²

We have to find height of the building.

As per work energy theorem, work done is always equal to the change in energy of body.

Therefore, W = ∆U

where, W = work done and ∆U = change in potential energy.

Potential energy of body of mass m at height h is given by; U = mgh

We know that, potential energy at ground is taken zero.

→ W = mgh - 0

→ 5000 = 20 × 10 × h

→ 5000 = 200h

→ h = 50/2

→ h = 25m

Height of the building = 25m

Answer 2

Correct Question :-

If 5KJ of work is done in lifting 20kg of water to the top of a building , calculate the height of the building ( take g = 10 m /s)

$\Large{\underline{\underline{\bf{Answer :-}}}}$

$\large{ \boxed{\bf Height \: = 25 \:\:m }}$

Given :-

✏ Mass of water - 20 kg

✏ Work done = 5kJ

✒ 1kJ = 1000J

$\therefore\underline{\boxed{\textsf{5kJ = {\textbf{5000J}}}}} \qquad\qquad \bigg\lgroup\bold{Converting \ into \ joules} \bigg\rgroup$

✒ Mass of water = 20kg

✒ Value of g = 10m/s²

Have to find out :+

✒ Height of the building

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❥ Now ,

As according to the Question

We know that ,

✒ Work done = change in energy

✏ ( W = ∆U )

where, W = work done and ∆U = change in potential energy.

✏ Potential energy of body of mass m { U = mgh }

Potential Energy at ground remains zero always

Now ,

↠ W = mgh - 0

↠ 5000 = 20 × 10 × h

↠ 5000 = 200h

↠ h = 50/2

↠ h = 25m

$\large{ \boxed{\bf Height \: = 25 \:\:m }}$

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