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

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Homework Help Overview

The discussion revolves around calculating the work done by a force of 152N over a distance of 13m at an angle of 33°. The original poster expresses confusion regarding the rounding of the calculated work from 1657.2J to 1660J, seeking clarification on the reasoning behind this rounding in the context of significant figures.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of work using the formula W = F * Δs * Cosθ and question the significance of rounding the result based on significant figures. There is an exploration of how the number of significant figures in the given values affects the final answer.

Discussion Status

Participants have provided insights into the concept of significant figures and how they apply to the rounding of the final answer. There is an ongoing exploration of the implications of significant figures in the context of the problem, with some participants clarifying the rules without reaching a definitive consensus on the interpretation of the distance's significant figures.

Contextual Notes

There is a discussion about the significant figures of the force and distance values used in the calculation, with some participants noting potential discrepancies in the number of significant figures for each measurement.

DaDoctor
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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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