Power Needed!!

Classical Mechanics Level 2

A fully loaded, slow-moving freight elevator has a cab with total mass of 1200 kg, which is required to travel upward 54 m in 3.0 min, starting and ending at rest. The elevator's counter-weight has a mass of only 950 kg, and so the elevator motor must help. What average power (in Watts) must be generated by the motor to move the elevator?

Image credit: Gensler


The answer is 7.4E+2.

Bhaskar De by Bhaskar De , 23

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

8 solutions

Ankit Singh
Mar 8, 2014

As the counter-weight can lift 950 kg, the motor has to lift up the remaining 250 kg. Now, W = F.s So, W = 250 . 9.8 . 54J Time t = 3 . 60s = 180s Therefore, Power, P = 250 . 9.8 . 54/180 = 735J

5 Helpful 0 Interesting 0 Brilliant 0 Confused

P=W/t and W=mgh

Krishna Das - 7 years, 3 months ago

iam not under standing why we subtract << i think that 1200+950

Prince Eldonia - 7 years, 2 months ago

Log in to reply

power = work done /time taken SI unit = JS^-1 so we will use gravitational potential energy as the lift goes up and down not left or right(accelerating) so gpe=mgh mass x gravitational pull x height = 950kg x 10 ms^-2 x 54m =513000J use it in power's formula which is p=w/t= 513000J/180s=2850JS-1 which is = to 2850watts (W)

Adeel Adam - 7 years, 2 months ago

power = work done /time taken SI unit = JS^-1 so we will use gravitational potential energy as the lift goes up and down not left or right(accelerating) so gpe=mgh mass x gravitational pull x height = 950kg x 10 ms^-2 x 54m =513000J use it in power's formula which is p=w/t= 513000J/180s=2850JS-1 which is = to 2850watts (W)

Adeel Adam - 7 years, 2 months ago

Power is not measured in Joules. Fix it.

Bernardo Sulzbach - 6 years, 11 months ago
Shadman Saif
Mar 9, 2014

The mass of the counter weight cancels out most of the mass of the lift. Thus the motor has to lift only (1200-950=250)Kg. Now, P=W/t =>P=mgh/t =>P=(250x9.81x54)/(3x60)
=>P=735.75

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Souradeep Guha
Mar 17, 2014

P=Force * avg velocity

Avg velocity = total disp/total time = 54/180

so, power = (1200-950=250)*9.81 * (54/180)=735.75

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Kahsay Merkeb
Mar 15, 2014

First calculate the force which would be mg,1200-950=250 so 250 9.5m/s^2=2450 Now, P=W/t =mgh/time finally 2450 54/180=735Joules

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Amit Kumar singh
Mar 20, 2014

THE POTENTIAL ENERGY OF THE CART TA THE TOP IS mgh =(1200)(9.81)(54) J. THIS ENERGY COMES FROM THE POTENTIAL ENERGY OF THE COUNTER WEIGHT i.e( 950)(9.81)(54) J PLUS THE ENERGY PROVIDED BY THE MOTOR. HENCE THE ENERGY PROVIDED BY THE MOTOR IS (250)(9.81)(54) J. NOW POWER IN WATTS IS ENERGY DISSIPATED PER SECOND. HENCE POWER = (250)(9.81)(54)/180 = 735.75 WATTS

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Mg=mg+F,where M=mass of cab,m=counter mass and f=force provided by motor,g=gravity. v=s/t. where s is the upward distance and t is the time taken to reach upward(in secs) P=Fvwhere P is power

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Ahmed Safiullah
Mar 11, 2014

Here, Work done= mgh, where m= (1200-950)kg, g=9.8 m/s*s , h=54m, Time (t)=3 min Power = Work done / t= 735 J

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Ian Valondo
Mar 11, 2014

Given: Mass1=1200 kg; Mass2= 950 kg; Distance=54 m; Time= 3 min = 3(60) =180 secs

Formula: Power=(Force*Distance)/Time

Solution: Power1 is power required to move the freight up to final position; Power2 is power supplied by the counterweight; Power3 is power supplied by the motor

  1. Power1=Power2 +Power3
  2. [(Mass1)(9.806)(Distance)/(Time)]=[(Mass2)(9.806)(Distance)/(Time)] + Power3
  3. [(1200)(9.806)(54)/(180)]=[(950)(9.806)(54)/(180)] + Power3
  4. 3530.16=2794.71 + Power3
  5. Power3=735.45
0 Helpful 0 Interesting 0 Brilliant 0 Confused

i think that is right

power = work done /time taken SI unit = JS^-1 so we will use gravitational potential energy as the lift goes up and down not left or right(accelerating) so gpe=mgh mass x gravitational pull x height = 950kg x 10 ms^-2 x 54m =513000J use it in power's formula which is p=w/t= 513000J/180s=2850JS-1 which is = to 2850watts (W)

Adeel Adam - 7 years, 2 months ago

I can understand your method..

Hafizh Ahsan Permana - 7 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...