# Math Help - Amount of work to raise bucket

1. ## Amount of work to raise bucket

A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 55 meters above the ground. This takes 12 minutes, during which time 5 kg of water drips out at a steady rate through a hole in the bottom. Find the work needed to raise the bucket to the platform. (Use .)

2. Originally Posted by amiv4
A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 55 meters above the ground. This takes 12 minutes, during which time 5 kg of water drips out at a steady rate through a hole in the bottom. Find the work needed to raise the bucket to the platform. (Use .)
remember, the work is given by:

$W = \int_a^b F(x)~dx$

where $F(x)$ is the force required to do the job of lifting an object from a height $x = a$ to $x = b$

obviously $a = 0$ and $b = 55$. the hard part is to find $F(x)$.

so lets think through this. it takes 20(9.81) N = 196.2 N of force to lift a 20kg bucket. but this force is changing, since water is leaking out. by the time we get to the top, 5 kg of water leaked out. the trip took 12 minutes, so that means the water is leaking at 5/720 = 1/44 kg/s. in other words, every second, the force we are lifting decreases by 1/44 kg = 0.068125 N.

now, can you tell me what $F(x)$ is?

3. 196.2x-(1/44)x dx ?????

4. Originally Posted by amiv4
196.2x-(1/44)x dx ?????
oh, i forgot to say what x was, though you may have figured it out. x is the distance from the ground to the bucket. that is, it represents the distance we have already lifted the bucket. you have to say that.

so no, this is not right.

here's another hint. when the bucket is half way up, you would want 2.5 kg of water to leak out, so the force would be 171.675. so whatever your formula is, it must equal 171.675 when x = 27.5

5. that just confused me more

6. Originally Posted by amiv4
that just confused me more
ok, lets do this from scratch. maybe you can apply the concepts here to the other problem

we have a 20kg bucket leaking 5/720 = 1/144 kg/s.

we travel 55/720 = 11/144 m/s. so that means we are losing 1/144 kg every 11/144 meters, or in other words, we are losing 1/11 kg/m

thus, after lifting x meters, the mass we are lifting is $20 - \frac 1{11} \cdot x$ kg.

now, force = mass*gravity, so that

$F(x) = 9.81 \bigg( 20 - \frac x{11} \bigg)$

so that the work done is $W = 9.81 \int_0^{55}\bigg( 20 - \frac x{11} \bigg)~dx$

7. k so it is 9442.125 J ????

8. Originally Posted by amiv4
k so it is 9442.125 J ????
why do you always ask questions with so many question marks?

yup, that's what i got.

the important thing though, is did you get the concept? you realize why i did what i did? i guess we will find out. i am waiting on your solution for your other work problem