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Sagot :
Explanation:
Given,
- m = 100 kg
- g = 10 N/kg¹
- h = 60 m
- t = 20 s
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:
- a) Work done by the pump
We know that,
[tex] \rm \: Work \: done = Force * Distance \: moved[/tex]
- f = 100 kg * 10N/kg
- d = 60 m
[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]
- b) Potential energy stored in the water
[tex] \rm \: P.E = m \cdot g \cdot h[/tex]
- m = 100 kg
- g = 10N/kg
- h = 60
[tex] \rm \: P.E =100 \:kg \: \times \cfrac{10 \: N}{kg} \times 60[/tex]
[tex] \boxed{\rm \: P.E =60000 \: joule}[/tex]
- same condition here as well, N×M = J
- c) Power of the Pump
[tex] \rm \: P = W/T[/tex]
- where P = Power; W = Work done & T = Time taken
- As we got the value of work done on question (a),& ATQ time taken is 20 S.
[tex] \rm \: P = \cfrac{60000 \: joule}{20 \: seconds} =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}[/tex]
- d) Power rating of the pump = 3 kW
Assumption: The pump is 100% efficient & works well.
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