Explanation:
Given,
To Find:
a) Work done by the pump
b) Potential energy stored in the water
c)Power spent by the pump
d)Power rating of the pump.
Solution:
We know that,
[tex] \rm \: Work \: done = Force * Distance \: moved[/tex]
[tex] \rm \: Work\; Done =(100 \: kg \times \cfrac{10N}{kg} ) \times 60 \: m[/tex]
[tex] \rm \: Work\; Done =1000 \times 60 \: joule[/tex]
[tex] \boxed{\rm \: Work\; Done =60000 \: joule}[/tex]
[The unit'll be joule since N×M = J]
[tex] \rm \: P.E = m \cdot g \cdot h[/tex]
[tex] \rm \: P.E =100 \:kg \: \times \cfrac{10 \: N}{kg} \times 60[/tex]
[tex] \boxed{\rm \: P.E =60000 \: joule}[/tex]
[tex] \rm \: P = W/T[/tex]
[tex] \rm \: P = \cfrac{60000 \: joule}{20 \: seconds} =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}[/tex]
Assumption: The pump is 100% efficient & works well.