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A pump is used to lift 100 KG of water from a wel 60 m deep,in 20 S If force of gravity on 1 KG is 10 N,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

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.