A Pretty Basic Work/Energy Question

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The discussion revolves around calculating the work done by a man pushing a 30.0 kg crate over 4.50 meters at constant velocity, considering a coefficient of kinetic friction of 0.250. The formula used is W = F*d, where the force is derived from the frictional force acting against the man's push. The calculated work is 331 joules, while the textbook states 333 joules, leading to questions about potential errors in the textbook or rounding discrepancies. The consensus suggests that the method used is correct, and the difference may stem from a typo or rounding in the textbook. The discussion highlights the importance of verifying calculations against authoritative sources.
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Homework Statement



A man pushes on a 30.0kg crate 4.50 meters at a constant velocity. The floor has a coefficient of kinetic friction of 0.250. How much work does the man's force do on the crate.

Homework Equations



W = F*d, ƩFx = F(push) - F(friction) = 0

The Attempt at a Solution



W = F*d, where F = 0.25*30.0kg*9.81m/s^2*4.50m

W = 331 joules

My book has 333 joules, is my book wrong? If not, where are they getting the extra joules?
 
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I don't see anything wrong with your method. Maybe the book answer is a typo or a rounding error of some kind?
 
cepheid said:
I don't see anything wrong with your method. Maybe the book answer is a typo or a rounding error of some kind?

Thanks for your opinion.
 

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