A 1.0 x 10^3 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of 4.0 x 10^3 N s the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.0 m/s?
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To find the minimum power the motor must deliver to lift the fully loaded elevator at a constant speed of 3.0 m/s, we need to consider the various forces acting on the elevator. We'll use the following formula to calculate power:
Power (P) = Force (F) × Velocity (v)
First, let's calculate the net force acting on the elevator. The weight of the elevator is given by:
Weight (W) = mass (m) × acceleration due to gravity (g)
Given that the mass of the elevator is 1.0 x 10^3 kg, and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the weight:
W = (1.0 x 10^3 kg) × (9.8 m/s²) = 9.8 x 10^3 N
Next, we need to calculate the force exerted by the motor to lift the fully loaded elevator. This force consists of two components: the weight of the elevator and the frictional force opposing its motion:
Force exerted by motor (F_exerted) = Weight + Frictional force
F_exerted = 9.8 x 10^3 N + 4.0 x 10^3 N = 13.8 x 10^3 N
Finally, we can calculate the power using the formula mentioned earlier:
Power = Force exerted × Velocity
Power = (13.8 x 10^3 N) × (3.0 m/s)
Now, let's convert the power from watts to kilowatts:
Power (in kilowatts) = (13.8 x 10^3 N × 3.0 m/s) / 1000
Power (in kilowatts) ≈ 41.4 kW
Therefore, the minimum power the motor must deliver to lift the fully loaded elevator at a constant speed of 3.0 m/s is approximately 41.4 kW.
To find the minimum power delivered by the motor to lift the fully loaded elevator at a constant speed of 3.0 m/s, we need to consider the work done against gravity and the work done against friction.
1. Work done against gravity:
The weight of the fully loaded elevator is given by:
weight = mass × acceleration due to gravity
weight = (800.0 kg + 1.0 x 10^3 kg) × 9.8 m/s^2
weight = 1.8 x 10^3 kg × 9.8 m/s^2
weight = 1.764 x 10^4 N
The work done against gravity is given by:
work_gravity = weight × distance
Since the elevator is moving at a constant speed, the net force is zero and the work done against gravity is equal to the work done against friction:
work_gravity = work_friction
2. Work done against friction:
The work done against friction is given by:
work_friction = force_friction × distance
work_friction = (4.0 x 10^3 N) × distance
Since both the distance traveled against gravity and against friction are the same (as they are equal), we can set these two equations equal to each other:
work_gravity = weight × distance
work_gravity = (1.764 x 10^4 N) × distance
Setting this equal to the work done against friction:
(1.764 x 10^4 N) × distance = (4.0 x 10^3 N) × distance
Now we can solve for the distance:
(1.764 x 10^4 N) × distance = (4.0 x 10^3 N) × distance
1.764 x 10^4 N = 4.0 x 10^3 N
1.764 x 10^4/4.0 x 10^3 = distance
4.41 m = distance
3. Power:
The power is given by:
power = work/time
In this case, the work done is equal to the work against friction and the time is equal to the distance divided by the speed:
power = work_friction / time
power = (4.0 x 10^3 N) × (4.41 m) / (3.0 m/s)
Calculating the power:
power = (1.764 x 10^4 N × m) / s
power = 1.764 x 10^4 J/s
power = 1.764 x 10^4 W
To convert watts to kilowatts:
power = 1.764 x 10^4 W = 17.64 kW
Therefore, the minimum power, in kilowatts, that the motor must deliver to lift the fully loaded elevator at a constant speed of 3.0 m/s is 17.64 kW.
F=(m+M) g +F(fr)= (1000+800)9,8 + 4000 = 5440 N
P=Fv = 5440•3=16320 W