Finding Average Power: Elevator Acceleration and Cruising Speed Calculation

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To find the average power of a 650 kg elevator accelerating upward, first calculate the work done using the formula for kinetic energy, considering the final speed of 1.75 m/s after 3 seconds of acceleration. The work done on the elevator results in an increase in kinetic energy, which can be determined by the equation KE = 0.5 * m * v^2. Since the elevator experiences constant acceleration, the average velocity can also be calculated, simplifying the power calculation. The discussion emphasizes understanding the relationship between work, energy transfer, and the types of energy involved in the elevator's motion. Ultimately, the average power can be derived from the total work done over the time interval.
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A 650 kg elevator starts from rest. It goes upward for 3 seconds with constant acceleration until it reaches its cruising speed 1.75 m/s. Find average power...

I'm not really sure where to begin I am basically having trouble finding work for this problem, does anyone know how I should find the work for this?
 
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Start with the definition of acceleration...
 
The effect of Work is Energy transfer ...

Does the elevator have more Energy, later?
What kinds of Energy will it have?

(alternative approach: Force is constant,
average velocity is easy to determine.)
 
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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