The mass of a bucket full of water is 15kg it is being pulled up from 15m deep well. Due to hole in the bucket 6kg water flows out of the bucket . The work done in drawing the bucket out of the well will be

Question

amistre64 · Answer

whats our formula for work?

anonymous · Answer

plz ans in detail

amistre64 · Answer

no, this site is a study site, not an answer in detail site.  your participation is required for a solution to be approached.  in other words, you have to do some of the work and thinking.

anonymous · Answer

i think you also dont know  u are just hitting a trial

amistre64 · Answer

so are you asking a valid question, or are you just wasting our time? spamming the site?

if its a valid question, then we will need to use the formula for 'work'  and theres part of your question that is a bit vague that needs to be cleared up.

anonymous · Answer

it is a valid question, i mean u shld explain the variable part and upper and lower limit of the intgral

amistre64 · Answer

the variable part is the force portion of the work formula

the bucket is losing mass along the way such that the total mass lost is defined.

setting up a representative section of it tends to help iron out the details.

amistre64 · Answer

yet its still a good starting point to define the general formulas needed; therefore, what is the formula for work? in general that is.

anonymous · Answer

$$\int\limits_{?}^{?} F.ds$$

amistre64 · Answer

good, now Force is a measure of weight, so mass * gravity = F

mass is changing (15-m)

now the vagueness, can we assume the rate of change in mass is constant?  

6kg per 15 meters

anonymous · Answer

lower limit is 0 and upper is 15

amistre64 · Answer

yes, the lower and upper limits are 0 and 15

how dies the mass change with respect to the distance traveled?

anonymous · Answer

i m not getting that part plz ans

amistre64 · Answer

how much have we lost over 15 meters of travel?

anonymous · Answer

6kg

amistre64 · Answer

then would you agree that if this is a constant rate of change .. then we can assume

m = 6/15 s

when s=0, mass loss = 0
when s=15, mass lost is 6

sounds fair?

amistre64 · Answer

$$W=\int_{0}^{15}F~ds$$
$$W=\int_{0}^{15}mg~ds$$
$$W=\int_{0}^{15}9.8(15-\frac{6}{15}s)~ds$$

anonymous · Answer

ya i got 1764

anonymous · Answer

and if g=10 thn ans is 1800 J

anonymous · Answer

thnx

amistre64 · Answer

good job

anonymous · Answer

are u a physics lect

amistre64 · Answer

no, im a mathematician.  i can do some simple physics calcuations is all :)

anonymous · Answer

nice