How Much Power is Needed to Move an Elevator at 4.50 m/s?

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To determine the power required to move an elevator cab with a mass of 4600 kg and a maximum load of 1800 kg at a speed of 4.50 m/s, the total mass is 6400 kg. The power calculation involves using the formula P = F * v, where F is the force due to gravity (6400 kg * 9.8 m/s²) and v is the velocity (4.5 m/s). This results in a power requirement of 282240 Watts, which can be rounded to approximately 280 kW. The discussion clarifies the relationship between distance and time, confirming that D/T equals velocity. The final answer for the power needed is approximately 280 kW.
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Homework Statement


An elevator cab has a mass of 4600 kg and can carry a maximum load of 1800 kg. If the cab is moving upward at full load at 4.50 m/s, what power is required of the force moving the cab to maintain that speed?

Homework Equations


P = W/T
W = F*D
F = M*A

The Attempt at a Solution


P = W/T
P = (F*D)/T
P = ((M*A)*D)/T
P = ((6400*9.8)*D)/T
I don't know what I should enter for D and T
Help, please?
 
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D/T is equal to what ? (using kinematic equations ).
 
Distance and Time
 
Distance divided by time equals _____? (Fill in the blank)
 
oh velocity
 
so the answer would be 6400*9.8*4.5 = 282240 Watts?
 
GalacticSnipes said:
oh velocity
Well, D/T is average or constant speed , which is given.
 
constant
 
GalacticSnipes said:
so the answer would be 6400*9.8*4.5 = 282240 Watts?
GalacticSnipes said:
so the answer would be 6400*9.8*4.5 = 282240 Watts?
Yes. You should round it off to say 280 kW.
 
  • #10
But 282.240 kW is the non rounded answer?
 
  • #11
GalacticSnipes said:
But 282.240 kW is the non rounded answer?
Yes.
 
  • #12
ok, thx
 
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