[SOLVED] a woman carries a bucket to the top of a 40 m tower...


by iamkristing
Tags: bucket, carries, solved, tower, woman
iamkristing
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#1
Mar13-08, 02:54 PM
P: 33
1. The problem statement, all variables and given/known data

a woman carries a bucket of water to the top of a 40 m tower at a constant vertical speed. the bucket has a mass of 10 kg and initially contains 30 kg of water, but it leaks at a constand rate and only 10 kg of water are in the bucket at the top of the tower.

a) write an expression for the mass of the bucket plus water as a function of the height (y) climbed

b) find the world done by the woman on the bucket

2. Relevant equations

W= kf-ki

kf+uf=ki+ui

k=1/2mv^2

u=mgh

3. The attempt at a solution

a) Now i assumed that ui=0 since there is no height and kf=0 since the woman stops at the top of the hill.

now that gave me (m-20 kg)gh=1/2mv^2
or (m-20)/m=v^2/2gh

Now my problem here is I wasn't quite sure what the question was asking. I believe we already have the mass of the bucket and the water so I think the question just wants an equation with variables?

b) I used the equation above to solve for V, where m=40 h=40
V then =19.8 and i plugged that value into

W=1/2mv^2

=1/2(40)(19.8)^2=7841 J

now it seems this answer is nowhere near correct...any help would be appreciated!
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Hootenanny
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#2
Mar13-08, 03:16 PM
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Quote Quote by iamkristing View Post
3. The attempt at a solution

a) Now i assumed that ui=0 since there is no height and kf=0 since the woman stops at the top of the hill.

now that gave me (m-20 kg)gh=1/2mv^2
or (m-20)/m=v^2/2gh

Now my problem here is I wasn't quite sure what the question was asking. I believe we already have the mass of the bucket and the water so I think the question just wants an equation with variables?
The question is asking you how the mass of the bucket (and the water in the bucket) changes with respect to height. So you want to express mass (m) as a function of height (y), m = m(y).
lavalamp
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#3
Mar13-08, 03:23 PM
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For the first part, you know that the vertical speed is constant, and the leakage is constant. Also you were given start and end values for the mass of the water so you can calculate how much water was lost in total. Therefore you can deduce that when half of the total lost water was lost, she was half way up, and when 3/4 of the total lost water was lost she was 3/4 up etc. So you can consider height and mass of water loss as directly proportional to one another:
[tex]h \propto m[/tex]

Therefore:
[tex]h = km[/tex]

Where k is some constant you can calculate from known values of height and water loss. That will get you pretty close to your answer, but what you need to find is an equation for the mass of the water left plus the mass of the bucket, I'll let you see if you can take it from here though.

For the second part, you could just say that since the water loss is constant, you can use the average mass of the water in the bucket plus the mass of the bucket and plug it into the formula:
[tex]GPE = mgh[/tex]


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