Calculating Work Done by a Horse Pulling a Wagon with Friction on a Level Road

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The discussion focuses on calculating the work done by a horse pulling a 200 kg wagon over a distance of 80 km on a level road, with a coefficient of friction of 0.060. The work done is calculated using the formula W=Fd, resulting in a total work output of 14,739,200 Joules after accounting for frictional forces. The force of friction is determined to be 117.6 N, which is derived from the effective coefficient of friction multiplied by the normal force. The calculations are confirmed to be accurate and demonstrate the importance of clearly stating assumptions in physics problems.

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gdhillon
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:How much work did a horse do that pulled a 200 kg wagon 80 km along a level road if
the effective coefficient of friction was 0.060?




I used W=Fd to find the work (W=1960*80000=156800000) then I got the force of friction by ff=(coeff)Fn and got (.06)*1960=117.6. I then got the W(friction)=117.6(80000)=-9408000 and by subtracting the 2 Work answers I got 14739200J
 
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Great - did you have a question?
Note: you have an implicit assumption in the calculation that you should make explicit of this is a long-answer.
 
Sorry, stupid question but how did you know the Force was 1960 N for the 200kg?
 

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