Elastic Potential Energy Question

[SDPHD]

I must be extremely stupid or out of it to not understand this question right now.



I know the work done equals the area under the graph. I estimate 14 squares, but how do I convert this to Joules?

Thanks

[onur369]

The work done by the archer on the bow is equal to the area under the force–distance graph
from XY = 0.10 m to XY = 0.30 m.
This area = 7.2 J.

Also another tip for this type of question is using the trapezium method, assume it looks like a half trapezium.
Flip the graph 90degrees into the right direction and you will get a half trapezium.
(a+b)/2 x h
(30+50)/2 x h
40x0.2 = 8.
Which is approximate, hope this helps.

[SDPHD]

The work done by the archer on the bow is equal to the area under the force–distance graph
from XY = 0.10 m to XY = 0.30 m.
This area = 7.2 J.

Also another tip for this type of question is using the trapezium method, assume it looks like a half trapezium.
Flip the graph 90degrees into the right direction and you will get a half trapezium.
(a+b)/2 x h
(30+50)/2 x h
40x0.2 = 8.
Which is approximate, hope this helps.

I'm a complete idiot! I just realised that one square is equal to 0.5J as the length of the square is 0.05m and the height is 10N. :idiot2:

I never thought of using your trapezium method! I'll keep that in mind for future area under graph questions!

Thanks for the help!

[onur369]

all good, we learn from out mistakes. Rather do mistakes before its too late.

[schnappy]

That is the most mind fuckingly worded question I've ever read.

[SDPHD]

That is the most mind fuckingly worded question I've ever read.

LOL! :2funny:

Welcome to the wonderful world of Heinemann. :tickedoff: