Work - Acceleration of an elevator

In summary, the elevator with a mass of 1300 kg accelerates from 0 m/s to 1.5 m/s in 2 seconds and then continues with a constant velocity of 1.5 m/s. The work needed to accelerate the elevator is 21548.475 Joule, taking into account the change in kinetic energy, potential energy, and frictional force.
  • #1
tobias.mueller2
2
0

Homework Statement



An elevator with 1300 kg accrlerate from 0 m*s^-1 to 1.5 m*s^-1 in 2 s. After that, he drives further with a constant velocity of 1.5 m*s^-1. One must add a frictional force of 637,65 N.

The questions are:
1. How much work is needed to accelerate the elevator?
2. How much work is needed for the driving of the elevator with a constant velocity in dependence of the height?

Homework Equations



The Attempt at a Solution


1.
w=m*a*s+637,65 N*s
a=v*t^-1+9,81 m*s^-2=1.5 m*s^-1*(2 s)^-1+9,81 m*s^-2=10,56 m*s^-2
s=0,5*a*t^2=0,5*1.5 m*s^-1*(2 s)^-1*(2 s)^2=1.5 m
w=1300 kg*10,56 m*s^-2*1.5 m+637,65 N*1.5 m=21548.5 Joule

2.
w=m*a*s+637.65*s
w=1300 kg*9,81 m*s^-2*s+637.65*s

I guess 21548 Joule is a way too many for just the acceleration of an elevator, isn't it?
 
Last edited:
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  • #2
The equations are correct, but the acceleration derivation is incorrect.

a=v*t^-1+9,81 m*s^-2=1.5 m*s^-1*(2 s)^-1+9,81 m*s^-2=10,56 m*s^-2

The work equivalent to the change in kinetic energy. The elevator starts at rest (KE = 0) and accelerates to a new KE (0.5* m * v2).

The distance of 1.5m is correct, and the contribution of friction (637,65 N*1.5 m) is correct.

In the first part, does one need to consider the change in graviational potential energy, mgh or mgs, where s is the vertical distance traveled (= h)? This is considered in the second part.

Work/energy = force * distance, assuming constant force, otherwise if F = F(x), then one must integrate [itex]\int{F(x)}\,dx[/itex]
 
  • #3
Astronuc said:
The work equivalent to the change in kinetic energy. The elevator starts at rest (KE = 0) and accelerates to a new KE (0.5* m * v2).
Astronuc said:
need to consider the change in graviational potential energy, mgh or mgs,


w=∆e[kyn]+∆e[pot]+637,65 N*1.5 m=0.5*m*v2-0+m*g*s-0+637,65 N*1.5 m=0.5*1300 kg*1.5 m/s2+1300 kg*9.81 m/s2*1.5 m+637,65 N*1.5 m=21548,475 Joule

In the end, I get the same result. Do I made any fault?
 

1. What is the acceleration of an elevator?

The acceleration of an elevator is the rate at which the elevator's speed changes over time. It is typically measured in meters per second squared (m/s2).

2. How is the acceleration of an elevator calculated?

The acceleration of an elevator can be calculated by dividing the change in velocity by the change in time. This can be represented by the equation a = Δv/Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time.

3. What factors affect the acceleration of an elevator?

The acceleration of an elevator can be affected by several factors, including the weight of the elevator and its contents, the friction between the elevator and the shaft, and the strength of the motor.

4. How does the direction of motion affect the acceleration of an elevator?

The direction of motion can greatly affect the acceleration of an elevator. When the elevator is moving in the same direction as the acceleration, it will increase in speed. However, if the elevator is moving in the opposite direction of the acceleration, it will decrease in speed.

5. What is the significance of the acceleration of an elevator?

The acceleration of an elevator is important because it determines the comfort and safety of the passengers. A smooth and steady acceleration is ideal for a comfortable ride, while a sudden or jerky acceleration can be uncomfortable and potentially dangerous for passengers.

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