Calculating work done by gravity

[JustynSC]

Homework Statement

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?

Homework Equations

Work=Force*Distance (what thought to use)
W_ext=F_extdcosθ (what the answer key says to use.

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?***

 

[mfb]

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.

 

[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?

 

[mfb]

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

 

[JustynSC]

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

 

[CWatters]

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