# I understanding a concept about motor efficiency

• pizzamakeren
In summary, the power needed to move an elevator of weight 1250kg at a speed of 2m/s +/- 0,10m/s after 3 seconds is 825W. The motor efficiency (percentage of input power that is converted to output power) is 80%, so the required power increases by 25%.

#### pizzamakeren

Homework Statement
I need help solving a problem about motor efficiency and an elevator
Relevant Equations
Weight of elevator: 1250Kg. Speed after 3 seconds: 2m/s +/- 0,10m. Motor efficiency: 0,80
I have tried to solve a problem about the amount of watts a motor needs to recive if i want an elevator of 1250Kg to move at a speed of 2m/s +/- 0,10m/s after 3 seconds. The motor has an efficiency of 0,80. I've tried looking at my teachers uni recording, but it doesn't seem to explain this part very well. Could someone with knowledge about this try to teach me how i should handle this task? I know that the motor efficiency makes it so that i need more watts, as 20% of the energy put into the motor vanishes into other forms of energy.

pizzamakeren said:
Homework Statement:: I need help solving a problem about motor efficiency and an elevator
Relevant Equations:: Weight of elevator: 1250Kg. Speed after 3 seconds: 2m/s +/- 0,10m. Motor efficiency: 0,80

I have tried to solve a problem about the amount of watts a motor needs to recive if i want an elevator of 1250Kg to move at a speed of 2m/s +/- 0,10m/s after 3 seconds. The motor has an efficiency of 0,80. I've tried looking at my teachers uni recording, but it doesn't seem to explain this part very well. Could someone with knowledge about this try to teach me how i should handle this task? I know that the motor efficiency makes it so that i need more watts, as 20% of the energy put into the motor vanishes into other forms of energy.
Try to sort out your problem on the one hand, and the tools you have available on the other hand.m That solves three quarters of this forum's exercises. That's what the template is intended for. So:

Homework Statement:: how much power is needed to accelerate 1250 kg to 2 m/s in 3 seconds at 80 % efficiency

Relevant Equations:: \begin {align*} F &= ma\\ v &= v_0 + at \\ dW &= \vec F\cdot d\vec s \\ P&={dW\over dt} \\ P_{\rm out} &= \eta P_{\rm in} \end{align*}

Lnewqban
What is the meaning of +/- 0,10m?

It means it varies, Sometimes it's 1,9m/s and sometimes it's 2,1m/s

pizzamakeren said:
It means it varies, Sometimes it's 1,9m/s and sometimes it's 2,1m/s
Thank you.
Can you work some calculation based on post #2?

pizzamakeren said:
It means it varies, Sometimes it's 1,9m/s and sometimes it's 2,1m/s
Interesting. Will it never never never be 2.3 m/s or more ?

But since you probably want only one motor, you'll have to pick a single value .

BvU said:
Interesting. Will it never never never be 2.3 m/s or more ?

But since you probably want only one motor, you'll have to pick a single value .
it's going to stay at 2m/s after 3 seconds

Lnewqban said:
Thank you.
Can you work some calculation based on post #2?
F = 825
Im not sure what v0, Dw, ds, or dt means so i can't calculate the rest

@pizzamakeren, assuming the elevator is moving vertically, then as well as kinetic energy, you may be expected to consider the gain/loss in gravitational potential energy (depending on which direction elevator moves).

Also (being pedantic) there are 2 mistakes in the statement:
“Weight of elevator: 1250Kg“

Steve4Physics said:
@pizzamakeren, assuming the elevator is moving vertically, then as well as kinetic energy, you may be expected to consider the gain/loss in gravitational potential energy (depending on which direction elevator moves).

Also (being pedantic) there are 2 mistakes in the statement:
“Weight of elevator: 1250Kg“
Mass of elevator

pizzamakeren said:
Mass of elevator
The other mistake is Kg should be kg.
'k' means kilo (=1000).
'K' means kelvin (unit of temperature).