Finding how high a winch can lift a mass before the battery discharges

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Finding how high a "winch" can lift a mass before the battery discharges

Homework Statement



The EMF of the D cell battery is 1.5 V and the life is about 10 A/hr. 50% of the energy int he winch does actual work and the lifting mass is 50 kg. The winch has a load resistance variable of Rl.

Homework Equations


Total Power=VI-V^2/Rl

(1.5*10)-10^2/Rl Total power of the system since the question states that the winch's resistance is much greater than the battery's internal resistance. (Thus ignore battery's resistance.)


The Attempt at a Solution




After finding the total power of the system, I thought that convert the power into work and using mgh to find the max height of winch would work. But at this point, I do not understand how to convert power into work if I do not know the time. Am I to find the height in terms of hours? I've read the book and searched around the internet but the conversion between power and energy is confusing.


Thank You
 
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You are trying to fin the total ENERGY of the system.
power = V * I and energy = power * time = V * I * t

The capacity of the battery if 10 AHour (I*t) - you have written it down wrong.
 


O I see. That should help me quite a bit. Thank You
 


also, P is VI but don't I need to take into account the resistance of the winch it's self? Or does the 50% efficiency take into account of the resistance?
 


Yes the 50% efficency includes everything - the question is much simpler than you are imagining
 


so in the next problem, we have to assume that the 50% doesn't apply. I now have to find the Rl which would maximize the energy. ..

This is where I would apply the total energy-energy due to winch correct?

My question here is that how do I find energy loss due to the winch? Can i use E=I^2Rt? But I'm confused since I is squared. can I assume that 10 A*hour is just I also? so that energy is 10^2*Rl*t and power is 10^2Rl?