Work and Potential Energy Problem

AI Thread Summary
The problem involves calculating the weight of a load raised by a 140 Watt motor with 70% efficiency over a height of 5 meters in 5 seconds. The motor's effective output is determined to be 98 Watts, leading to a potential energy of 490 Joules. Using the formula PE = mgH, the calculated mass comes out to 10 kilograms. However, the expected answer is 110 kilograms, suggesting a discrepancy. The discussion concludes with a recommendation to consult the professor for clarification on the solution.
bengaltiger14
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Homework Statement



A 140 Watt motor (70% efficient) is available to raise a load 5 meters into the air. If the task takes 5 seconds to complete, how heavy was the load in kilograms?


I first determine the motot output: 140W * 70% == 98 Watts

I then multiply 98 Watts by 5 seconds to get PE = 490Joules

Now, setting 490Joules = mgH and solving for m, I get 10 kilograms

The answer is supposed to be 110kg according to my professor but I cannot get to that number. What am I doing wrong?

Thanks
 
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Your answer looks good to me. I don't think you did anything wrong. (Ask your professor to show her solution.)
 
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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