A pump is used to lift $500$ kg of water from a depth of $80$ m in $10$ s. Calculate the power rating of the pump if its efficient is $40 %$ . (Take $g = 10$ m $s^{- 2}$ ).

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Solution

$P o w e r = w o r k / t i m e = m g h / t i m e = (500 \times 10 \times 80) / 10 = 40000 W = 40 K W$

$S i n c e p u m p e f f i c i e n c y = 40$ = $(40 K J \times 40) / 100 = 16 K W$

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Similar questions

500 k g

80 m

10 s

40 %

A pump is used to lift 500 kg of water from a depth of 80 m in 10 s.Calculate:

(a) the work done by the pump,

(b) the power at which the pump works, and

500

80

10

500 k g

80 m

10 s

g = 10 m s^{- 2}

500

80

10

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