Find Solution to Ship Fuel Problem: 1.0 X 10^5 N Drag

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To solve the problem of how far a nuclear-powered ship can travel per kilogram of fuel with a drag force of 1.0 X 10^5 N, one must calculate the work done against the drag and the energy produced from the fission of uranium. The energy released per fission event is 208 MeV, and with enriched uranium containing 1.7% of 235U, the total energy per kilogram of fuel can be determined. Considering the engine's efficiency of 20%, the effective energy available for propulsion is calculated. Ultimately, this leads to the conclusion that the ship can travel approximately 2.9 X 10^3 km (1800 miles) per kilogram of fuel. Understanding the relationship between work, energy, and efficiency is crucial in reaching this solution.
laydico
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I need help finding the solution to this problem...

Suppose that the waterf exerts an average frictional drag of 1.0 X 10^5 N on a nuclear-powered ship. How far can the ship travel per kilogram of fuel if the fuel consists of enriched uranium containing 1.7% of the fissionable isotope 235U and the ship's engine has an efficieny of 20% (assume 208 MeV is released per fission event.)

I know the answer is 2.9 X 10^3 km (1800 miles) I just don't know how to get to it.
 
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One is suppose to show one's work before asking for assistance.

HINT: Thik about the definition of work done.

Regards,
~H
 
i figured it out
 
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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