Calculate Work Done by Force (152N): 1657.2J

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The work done by the force of 152N over a distance of 13m at an angle of 33° is calculated to be 1657.2J. The discussion centers on the need to round this value to 1660J, which is related to significant figures in the measurements used. Since the force has three significant figures and the distance has two, the final answer should reflect the least number of significant figures, which is two in this case. This rounding ensures that the precision of the result is consistent with the measurements provided. Understanding significant figures is crucial in scientific calculations to maintain accuracy.
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How much work is done by the force?
W= F * Δs * CosθW= (152N) * (13m) * Cos(33°)
W=1976 * Cos(33°)
W=1657.2

So I'm fairly confident I got this right since the answer is 1660J my question is Why would I have to round it up to 1660 to get it to Joules? I'm not really looking for help on the answer just an explanation as to WHY? Any help would be appreciated!
 
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It's already in joules. When you multiply force by distance,
W=F*d*cos(theta)
Joules = Newtons * meters * [no dimensions for trig]
 
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Engineer at UIC said:
It's already in joules. When you multiply force by distance,
W=F*d*cos(theta)
Joules = Newtons * meters * [no dimensions for trig]
ok but why the rounding to 1660 from 1657? That is kind of confusing to me so i really want to know to make sure I am not missing anything important.
 
DaDoctor said:
ok but why the rounding to 1660 from 1657? That is kind of confusing to me so i really want to know to make sure I am not missing anything important.

You can only answer with as many significant figures as you're given. 1657 is 4 significant figures, but your force only had three.
 
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Engineer at UIC said:
You can only answer with as many significant figures as you're given. 1657 is 4 significant figures, but your force only had three.
Oh... that makes a lot of sense! Thank you for your help!
 
Engineer at UIC said:
You can only answer with as many significant figures as you're given. 1657 is 4 significant figures, but your force only had three.
The force has three sig figs, but I believe the distance the box was moved (13 m) only has two.
 
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