Finding Work from the Derivative of Power

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SUMMARY

The discussion focuses on calculating the total work done by a machine delivering power at a decreasing rate defined by the equation P = P(o)*t(o)^2 / (t + t(o))^2. The user identifies the problem as potentially involving improper integrals and seeks guidance on how to approach the integration of power over time to determine the total work. The key takeaway is that integrating power over time yields the total work done, equating to energy delivered in Joules.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques
  • Familiarity with the concepts of power and work in physics
  • Knowledge of improper integrals and their applications
  • Basic understanding of units of measurement, specifically Joules and Watts
NEXT STEPS
  • Study integration techniques for improper integrals
  • Learn about the relationship between power, work, and energy in physics
  • Explore examples of integrating power functions over time
  • Review the concept of limits in calculus to better understand improper integrals
USEFUL FOR

Students in physics or engineering courses, educators teaching calculus and physics concepts, and anyone interested in understanding the relationship between power and work through integration.

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Homework Statement



A machine delivers power at a decreasing rate P = P(o)*t(o)^2 / (t + t(o))^2 , where P(o) and t(o) are constants. The machine starts at t = 0 and runs forever.

Find the amount of total work.

Homework Equations



P = w / t
W = f * d

P = P(o)*t(o)^2 / (t + t(o))^2

The Attempt at a Solution



I'm not really sure what to do here. My guess is that it is an improper (infinite) integral problem, but I'm just not sure how to really start. I tried to separate P(o) and t(o) from the equation since they are constants, but I couldn't get it work out.

Can anyone point me in the right direction?
 
Physics news on Phys.org
If you integrate power (Watts = Joules/sec) over time you will get Joules, which equates to the energy delivered. i.e., work. So do the integration :wink:
 

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