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

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To calculate the efficiency of a motor lifting a 0.050 kg block at a constant velocity of 0.40 m/s, the electrical power input is determined using the formula VIt, resulting in 0.255 W. The gravitational potential energy change is crucial, as it relates to the power required to lift the block against gravity. The efficiency can be calculated by comparing the useful power output (related to the potential energy change) to the electrical power input. Discussions highlight the importance of understanding the difference between kinetic and potential energy in this context. The final calculated efficiency of the motor is approximately 69.87%.
  • #31
malemdk said:
Rate of change of potential energy d(mgh)/dt =mg dh/dt = mg x v = Fx v = power needed to maintain the constant velocity, where from you can find efficiency if you know the input power
The fact that you used d(mgh)/dt confuses me, since we don't do that, and you put it on one line.
 
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  • #32
sorry
 
  • #33
TomHart said:
Just one minor issue. The voltage was 3.0 V.
yes
 
  • #34
IDK10 said:
The fact that you used d(mgh)/dt confuses me, since we don't do that, and you put it on one line.
Are you familiar with the equation Power = Force x Velocity?
 
  • #35
Yes , I designed many ,many machines using this equation .
it can be derived from Newtons second law
 
  • #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.
 
  • #37
ok
 

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