Question

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
​

Answer 1

Step-by-step 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,

`\rm \: Work \: done = Force * Distance \: moved`

• f = 100 kg * 10N/kg
• d = 60 m

`\rm \: Work\; Done =(100 \: kg * \cfrac{10N}{kg} ) * 60 \: m`

`\rm \: Work\; Done =1000 * 60 \: joule`

`\boxed{\rm \: Work\; Done =60000 \: joule}`

[The unit'll be joule since N×M = J]

• b) Potential energy stored in the water

`\rm \: P.E = m \cdot g \cdot h`

• m = 100 kg
• g = 10N/kg
• h = 60

`\rm \: P.E =100 \:kg \: * \cfrac{10 \: N}{kg} * 60`

`\boxed{\rm \: P.E =60000 \: joule}`

• same condition here as well, N×M = J
• c) Power of the Pump

`\rm \: P = W/T`

• 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.

`\rm \: P = \cfrac{60000 \: joule}{20 \: seconds} =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}`

• d) Power rating of the pump = 3 kW

Assumption: The pump is 100% efficient & works well.