# Homework Help: Calculating work done by gravity

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1. Nov 5, 2016

### JustynSC

1. The problem statement, all variables and given/known data
A 10kg bucket of water is lifted vertically 3.0m at a constant speed. How much work did gravity do on the bucket during this process?

2. Relevant equations
Work=Force*Distance (what thought to use)
Wext=Fextdcosθ (what the answer key says to use.

3. The attempt at a solution
My attempt got me as far as plugging in: Work=(9.8m/s2*10kg)*3.0m=(98N)*3.0kg=294J
***In the answer key they used 10m/s2 for gravity, but also had a negative answer of -300J. I assume that they used the negative as a direction, but is work a vector or scalar? Looking to understand why the way I did this does not yeild the same answer, and how the formula the keys says to use works?***

2. Nov 5, 2016

### Staff: Mentor

The force is in the opposite direction of motion, therefore work done by gravity is negative. You should change the sign of either the acceleration or the 3 meter distance, because they point in opposite directions.

3. Nov 5, 2016

### JustynSC

Okay thanks I was thinking something along those lines, but why would the key ask to solve using that other equation? It seems more complicated than the way I did it. does my method only work in limited scenarios and the other has a broader range of uses?

4. Nov 5, 2016

### Staff: Mentor

Well, the equation with the angle is more general, but as your problem is one-dimensional you don't need that.

5. Nov 5, 2016

### JustynSC

I see. In that case I will stick to my simple method :) Thanks!

6. Nov 5, 2016

### CWatters

In this case the angle is zero and Cos(0)=1 so both equations are the same.