Calculating Power for an Elevator with Maximum Load and Speed

[nns91]

Homework Statement

A 1200kg elevator driven by an electric motor can safely carry a maximum load of 800kg. What is the power provided by the motor when the elevator ascends with a full load at a speed of 2.3 m/s

Homework Equations

P=F*v

The Attempt at a Solution

So I use directly P=F*v=(m_elevator+m_people)*g*v

However I got the wrong answer. Does it involve any other formula here ?

 

[alphysicist]

Hi nns91,

Homework Statement

A 1200kg elevator driven by an electric motor can safely carry a maximum load of 800kg. What is the power provided by the motor when the elevator ascends with a full load at a speed of 2.3 m/s

Homework Equations

P=F*v

The Attempt at a Solution

So I use directly P=F*v=(m_elevator+m_people)*g*v

However I got the wrong answer. Does it involve any other formula here ?

What answer did you get?

 

[matt_crouch]

Dont you calculate the weight acting downwards so

w=mg

where the mass is 1200+800=2000kg

multiplied by g = 9.81ms^-2

so f downwards or the weight = 19620

then put it in

p=fv

so p= 19620 x 2.3

p= 45126 w


thats how i would do it don't know if I am right =]