- #1

ChiralSuperfields

- 920

- 106

- Homework Statement:
- Please see below

- Relevant Equations:
- Please see below

For this problem,

The answer is ##-4.70 kJ##. I am not sure what I am doing wrong.

My working is

## W = mgr\cos\theta ##

## W = mgr\cos150 ## (since angle between ##\vec g## and ##\vec r## is 150 degrees)

## W = -mgr\frac {\sqrt{3}}{2} ##

## W = -mgr\frac {\sqrt{3}}{2} ##

## W = (-80)(9.81)(12\sin60)(\frac {\sqrt{3}}{2}) ##

## W = -7063.2 J ##

Would some please be to offer some guidance?

Many thanks!

The answer is ##-4.70 kJ##. I am not sure what I am doing wrong.

My working is

## W = mgr\cos\theta ##

## W = mgr\cos150 ## (since angle between ##\vec g## and ##\vec r## is 150 degrees)

## W = -mgr\frac {\sqrt{3}}{2} ##

## W = -mgr\frac {\sqrt{3}}{2} ##

## W = (-80)(9.81)(12\sin60)(\frac {\sqrt{3}}{2}) ##

## W = -7063.2 J ##

Would some please be to offer some guidance?

Many thanks!