Calculating Electrical Energy Input for 70% Efficient Lift Motor

[Taylor_1989]

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?

 

[truesearch]

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

 

[flatmaster]

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?

 

[truesearch]

E = ?
just rearrange your equation...you have it !

 

[Norfonz]

Realize that if it is only 70% efficiency, more work is actually required than if it was 100% efficient.

 

[Taylor_1989]

Just to say thanks for everyone clearing it up for me big help.