How Much Work Do Each of the Forces Do on the Crate?

AI Thread Summary
The discussion centers on calculating the work done by three forces acting on a crate being dragged across the floor. The forces are specified with their magnitudes and angles, and the work is calculated using the formula w=Fscos(∂). Initial calculations yielded work values of 1804J, 1136J, and -2080J for the three forces. The user encountered issues with an online submission that indicated the answers were incorrect, later realizing the required unit was kilojoules instead of joules. This highlights the importance of unit conversion in physics problems.
DRC12
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Homework Statement


The three ropes shown in the bird's-eye view of the figure (Figure 1) are used to drag a crate 3.2 across the floor.How much work is done by each of the three forces? Then the picture is Force one is 600N 20o above the x-axis force two is 410N 30o below the x-axis and force 3 is 650N along the x-axis in the negative direction

Homework Equations


w=Fscos(∂)

The Attempt at a Solution


So I plugged the each of the numbers into the equation and got work from force one to be 600*3.2*cos20=1804J, the work from force two to be 410*3.2*cos30=1136 and work from force three equals 650*3.2*cos180=-2080
This seems really easy but when I plug it into the website it says it's wrong and gives no feedback I can't figure out what I'm doing wrong
 
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DRC12 said:
This seems really easy but when I plug it into the website it says it's wrong and gives no feedback I can't figure out what I'm doing wrong
Your work looks good to me. Sometimes they are fussy about the number of significant figures.
 
This site doesn't care about significant figures but it tends to be unclear about what it's looking for I just wanted to make sure I wasn't making any small mistakes
 
Just realized I was supposed to answer in kJ instead of J
 
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