Calculate Efficiency of Motor Lifting 0.050kg Block at 0.40m/s

In summary: It is true that at some time in the past, work was required to accelerate that weight up to its constant velocity. But that plays no role in the solution of this problem. What is relevant to the solution is the power required to maintain the increase in potential energy of the weight as it is being lifted at the specified rate. In other words, it is not the kinetic energy that matters; it is the increasing potential energy for which power must be supplied.In summary, to calculate the efficiency of the motor lifting a block with a mass of 0.050kg at a constant velocity of 0.40m/s, we need to consider the electrical power and kinetic energy equations. However, it is important to remember that
  • #36
malemdk said:
Yes , I designed many ,many machines using this equation .
I was asking @IDK10. He seemed to be having trouble understanding your solution.
 
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  • #37
ok
 

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