Maximizing Energy Usage: Swimming vs. Fast Food Meal

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A 68 kg person would need to swim at a fast crawl to expend the energy from a typical fast food meal, which provides 1350 Cal or 5670 kJ. With 25% of that energy being usable, the effective energy available for use is 1417.5 kJ. The swimmer's metabolic power output is 800 W, or 800 J per second. By setting up the equation, the time required to swim can be calculated as time equals energy divided by power. The final calculation indicates that the person would need approximately 1.77 hours to use all the energy from the meal.
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Homework Statement



For how long would a 68 kg person have to swim at a fast crawl to use all the energy available in a typical fast food meal of burger, fries, and a drink? Assume 25% efficiency.

Homework Equations



Fast food meal provides 1350 Cal or 5670 kJ
A 68 kg person swimming at a fast crawl uses 800 W of metabolic power.

The Attempt at a Solution



I'm having trouble just setting up the equation.
I know that Change in Thermal Energy + Change in Chemical Energy = 0 because it is an isolated system, but I don't know where to go from here, nor do I know what units I need each part to be in.
 
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Well, just think about this question for a second. This person intakes a certain amount of energy by eating the food. 25% of that energy is useable. The person then performs some activity that takes a certain amount of energy per second (or 800 J per second). You should be able to determine how long the person can swim just by giving the question some thought.
 
So would it be 5670*.25=800W* time?

Solving for time gives me 1.77 - but what units is this?
 
Op never mind, I needed to change kJ into Joules, then it gives me time in seconds. Whoops. :D
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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