# Efficiency question

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1. Feb 1, 2017

### IDK10

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Ek = 1/2 x mv2
Electrical power = VIt

3. 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?

2. Feb 1, 2017

### PeroK

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

3. Feb 1, 2017

### IDK10

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

4. Feb 1, 2017

### PeroK

Is the kinetic energy changing over time?

What if the block were moved horizontally at constant speed?

5. Feb 1, 2017

### IDK10

It is lifted at a constant velocity.

6. Feb 1, 2017

### PeroK

So, what's the change in Kinetic Energy?

7. Feb 1, 2017

0

8. Feb 1, 2017

### PeroK

So, what energy is changing?

9. Feb 1, 2017

### IDK10

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.

10. Feb 1, 2017

### PeroK

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

Why is it hard to lift something heavy?

11. Feb 1, 2017

### IDK10

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.

12. Feb 1, 2017

### PeroK

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.

13. Feb 1, 2017

### IDK10

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

14. Feb 1, 2017

### PeroK

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

15. Feb 1, 2017

### IDK10

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%

16. Feb 1, 2017

### malemdk

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

17. Feb 1, 2017

### PeroK

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.

18. Feb 1, 2017

### PeroK

This is nonsense!

19. Feb 1, 2017

### IDK10

Oh...

20. Feb 1, 2017

### IDK10

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%