Calculating Work and Power: Solving a Tractor Pull Problem

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SUMMARY

The discussion focuses on calculating work and power in a tractor pull scenario where a tractor exerts a constant force of 700N while pulling a wagon at a speed of 20.0 km/h. The work done by the tractor over 3.5 minutes is calculated to be 8.16 x 10^6 J, while the power output is determined to be 3888.8 watts. Key equations used include P=W/t and P=Fv, with conversions from km/h to m/s being essential for accurate calculations.

PREREQUISITES
  • Understanding of basic physics concepts such as work and power
  • Familiarity with the equations P=W/t and P=Fv
  • Ability to convert units, specifically from km/h to m/s
  • Knowledge of the relationship between force, velocity, and power
NEXT STEPS
  • Learn how to convert units between different measurement systems, focusing on speed conversions
  • Study the principles of work and energy in physics
  • Explore real-world applications of power calculations in mechanical systems
  • Investigate the impact of force and velocity on power output in various scenarios
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the calculations of work and power in mechanical systems, particularly in agricultural machinery contexts.

bob24
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Homework Statement



A tractor pulls a wagon with a constant force of 700N at a constant speed of 20.0 km/h.
a) How much work is done by the tractor in 3.5min
b) What is the tractor's power output?
ANSWERS: a)8.16 x 10^6 J//b) 3888.8 watts
2. Relevant equation
P=W/t or P=Fv



The Attempt at a Solution


P=Fv
P=700N x 20.0
P=14000J

Pt=W
14000W x 210seconds=W
W=2940000J

I've been trying for over an hour now and I just can't get the answer. Any help will be greatly appreciated.
 
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Convert km/h to m/s.
 
rl.bhat said:
Convert km/h to m/s.

I realized it about 5 seconds before you posted haha. Thank you very much!
 

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