IDK10 said:Homework Statement
A motor lifts a block of mass 0.050kg at a constant velocity of 0.40m s-1. The current in the motor ia 85mA and the potential difference across it is 3.0V. Calculate the efficiency of the motor
Homework Equations
Ek = 1/2 x mv2
Electrical power = VIt
The Attempt at a Solution
Ig to the electrical power as 0.255W, and kinetic energy of the block as 0.004J. But how do I get the power of the block, or the electrical energy of the motor?
Isn't efficiency useful (energy/power)/total (energy/power) x100PeroK said:Why do you think the kinetic energy of the block is important?
IDK10 said:Isn't efficiency useful (energy/power)/total (energy/power) x100
And kinetic energy is the useful energy.
It is lifted at a constant velocity.PeroK said:Is the kinetic energy changing over time?
What if the block were moved horizontally at constant speed?
IDK10 said:It is lifted at a constant velocity.
0PeroK said:So, what's the change in Kinetic Energy?
IDK10 said:0
Electrical to kinetic. Since I'm only given the block's mass, velocity, the current and voltage of the motor, I can't do power of the block, of the electrical energy of the motor.PeroK said:So, what energy is changing?
IDK10 said:Electrical to kinetic
How is it hard to lift 0.050kg (50g)? All I want to know is how to get the efficiency of a motor using the mass and velocity of a block, and the current and voltage of the motor.PeroK said:No. Kinetic energy is not changing. We have already established that.
Why is it hard to lift something heavy?
IDK10 said:How is it hard to lift 0.050kg (50g)? All I want to know is how to get the efficiency of a motor using the mass and velocity of a block, and the current and voltage of the motor.
But what do I do with it, there is no given height.PeroK said:I'll give you a big hint. At some stage, however, you are going to have to learn to think for yourself.
Hint: you're completely forgetting about gravity.
IDK10 said:But what do I do with it, there is no given height.
Do I do 0.4/9.81 to get the time in seconds, then times it by the current and voltage to get electrical energy. Then put that under kinetic energy, and times it by 100? It gives 38.5%PeroK said:I'm sorry, but you need to look at my last post. You need to think about the problem.
IDK10 said:Do I do 0.4/9.81 to get the time in seconds, then times it by the current and voltage to get electrical energy. Then put that under kinetic energy, and times it by 100? It gives 38.5%
malemdk said:Time to lift the mass is required
Kinetic energy /time gives you the power, then you compare the motor power with the calculated power to find efficiency of this system
Oh...PeroK said:I don't really understand why you are still trying to work with the kinetic energy. It's the gravitational potential energy of the block that is changing.
0.4/9.81 = 40/981sPeroK said:I don't really understand why you are still trying to work with the kinetic energy. It's the gravitational potential energy of the block that is changing.
IDK10 said:0.4/9.81 = 40/981s
0.4x40/981=16/981m
9.81x0.05x16/981=1/125J
(100/125)/(0.085x3x(40/981)) = 76.94% = 76.9%
From a curious observer, why did you divide 0.4 by 9.81?IDK10 said:0.4/9.81 = 40/981s
To accelerate a weight to 0.4m/s you need force, that force may be applied in 1sec or 100secPeroK said:This is nonsense!
In this problem you are not accelerating a weight. You are lifting a weight at a constant velocity.malemdk said:To accelerate a weight
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.malemdk said:To attain this velocity first you have to accelerate, if otherwise you reached the velocity with no time, which is physically not possible
0.05x9.8 x0.4/85 e-3*3 = 69.87%IDK10 said:Homework Statement
A motor lifts a block of mass 0.050kg at a constant velocity of 0.40m s-1. The current in the motor ia 85mA and the potential difference across it is 3.0V. Calculate the efficiency of the motor
Homework Equations
Ek = 1/2 x mv2
Electrical power = VIt
The Attempt at a Solution
Ig to the electrical power as 0.255W, and kinetic energy of the block as 0.004J. But how do I get the power of the block, or the electrical energy of the motor?
Just one minor issue. The voltage was 3.0 V.malemdk said:efficincy of your systemes = (0.05*9.8*0.4/85 e-3 x 3.3)x100 = 69.87%
The fact that you used d(mgh)/dt confuses me, since we don't do that, and you put it on one line.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
yesTomHart said:Just one minor issue. The voltage was 3.0 V.
Are you familiar with the equation Power = Force x Velocity?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.