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

• IDK10
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
IDK10

## 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?

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?

Why do you think the kinetic energy of the block is important?

PeroK said:
Why do you think the kinetic energy of the block is important?
Isn't efficiency useful (energy/power)/total (energy/power) x100
And kinetic energy is the useful energy.

IDK10 said:
Isn't efficiency useful (energy/power)/total (energy/power) x100
And kinetic energy is the useful energy.

Is the kinetic energy changing over time?

What if the block were moved horizontally at constant speed?

PeroK said:
Is the kinetic energy changing over time?

What if the block were moved horizontally at constant speed?
It is lifted at a constant velocity.

IDK10 said:
It is lifted at a constant velocity.

So, what's the change in Kinetic Energy?

PeroK said:
So, what's the change in Kinetic Energy?
0

IDK10 said:
0

So, what energy is changing?

PeroK said:
So, what energy is changing?
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.

IDK10 said:
Electrical to kinetic

No. Kinetic energy is not changing. We have already established that.

Why is it hard to lift something heavy?

PeroK said:
No. Kinetic energy is not changing. We have already established that.

Why is it hard to lift something heavy?
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.

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.

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.

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.
But what do I do with it, there is no given height.

IDK10 said:
But what do I do with it, there is no given height.

I'm sorry, but you need to look at my last post. You need to think about the problem.

PeroK said:
I'm sorry, but you need to look at my last post. You need to think about the problem.
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%

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

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%

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.

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

This is nonsense!

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.
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 = 40/981s
0.4x40/981=16/981m
9.81x0.05x16/981=1/125J
(100/125)/(0.085x3x(40/981)) = 76.94% = 76.9%

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%

You may have to ask your professor whether she/he would accept that. Personally, that would not be an acceptable solution as there is no explanntion of what you are doing.

IDK10 said:
0.4/9.81 = 40/981s
From a curious observer, why did you divide 0.4 by 9.81?

PeroK
PeroK said:
This is nonsense!
To accelerate a weight to 0.4m/s you need force, that force may be applied in 1sec or 100sec
Assume that weight attains 0.4m/s in 1 sec
The force =0.05x0.4/1=0.02N
Power = 0.02x0.04=0.0008W

malemdk said:
To accelerate a weight
In this problem you are not accelerating a weight. You are lifting a weight at a constant velocity.

PeroK
To attain this velocity first you have to accelerate, if otherwise you reached the velocity with no time, which is physically not possible

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
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.

PeroK
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

TomHart
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?
0.05x9.8 x0.4/85 e-3*3 = 69.87%

efficincy of your systemes = (0.05*9.8*0.4/85 e-3 x 3.3)x100 = 69.87%

malemdk said:
efficincy of your systemes = (0.05*9.8*0.4/85 e-3 x 3.3)x100 = 69.87%
Just one minor issue. The voltage was 3.0 V.

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.

sorry

TomHart said:
Just one minor issue. The voltage was 3.0 V.
yes

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?

Yes , I designed many ,many machines using this equation .
it can be derived from Newtons second law

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