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

**1. Homework Statement**

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. Homework 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!