Calculating Electrical Energy Input for 70% Efficient Lift Motor

  • Thread starter Thread starter Taylor_1989
  • Start date Start date
  • Tags Tags
    Efficiency Motor
Click For Summary
SUMMARY

The discussion focuses on calculating the electrical energy input required for a lift motor operating at 70% efficiency while performing 450,000 joules of work. The correct formula to determine the energy input (E) is E = 450,000 J / 0.7, resulting in an input of approximately 642,857 joules. Participants clarified that the energy input must account for the efficiency loss, emphasizing that 70% of the total energy consumed is utilized for lifting. This understanding is crucial for accurate energy calculations in mechanical systems.

PREREQUISITES
  • Understanding of basic physics concepts related to work and energy
  • Familiarity with efficiency calculations in mechanical systems
  • Knowledge of algebraic manipulation for solving equations
  • Basic understanding of electrical energy consumption
NEXT STEPS
  • Research electrical energy efficiency in motors and how it affects performance
  • Learn about energy conservation principles in mechanical systems
  • Explore advanced calculations for energy input in varying efficiency scenarios
  • Study the implications of efficiency on operational costs in lift systems
USEFUL FOR

Engineers, physics students, and anyone involved in the design or analysis of lift systems and motor efficiency calculations will benefit from this discussion.

Taylor_1989
Messages
400
Reaction score
14

Homework Statement


A lift motor is 70% efficient. It dose 450 000j of work lifting a load to the top floor of a building.



Homework Equations


What electrical energy input is required to do this?


The Attempt at a Solution



I first did output/input * 100 I transposed the formulae

so I did 450 000* 0.7

which gave the wrong answer. When I looked at the answer is say this is the correct way is 450 000 * 100/70

Am I right in saying that when you do it this you are calculating what it is at 100% efficiency and then calculating the difference?
 
Physics news on Phys.org
70% of the energy input is 450000J, so 1% is 450000/70 = 6429J
So 100% of the input energy is 100 x 6429 = 642900J
You had it 'upside down'...do you see
 
Last edited:
Write it out as an equation. Of the total electrical energy consumed, 70% of it went into lifting the weight.

E * 70% = 450,000 J

Where E is the unknown energy input.

How do you solve for E?
 
E = ?
just rearrange your equation...you have it !
 
Realize that if it is only 70% efficiency, more work is actually required than if it was 100% efficient.
 
Just to say thanks for everyone clearing it up for me big help.
 

Similar threads

Replies
36
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
17
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
8K
Replies
11
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K