1. An elevator shown below filled with passengers has a mass of 1650 kg. The elevator does motions (a)through (c) in succession.megFor each of the parts below draw a free body diagram of the elevator in your notebook for each ofthe parts (a) to (c). Draw the acceleration and velocity vectors in the boxes. For each part, are thevectors for tension in the string and weight of the elevator of equal lengths or unequal lengths.y1.XMEma m(a) The elevator accelerates upward from rest at a rate of 0.75s²(i) Newton's Second Law in the y-direction can be written as:Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards,pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration.xmeaΣF₁=T-meg=1(ii) Calculate the tension in the cable supporting the elevator.T= 17407.5N(iii) How high has the elevator moved during this time?Ay= 0.79(iv) Calculate the velocity of the elevator after this time.v(t = 1.45 s) = 1.09(b) The elevator continues upward at constant velocity for 8.1 s.(i) Newton's Law in the y-direction can be written as:Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards,pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration.mmSxmeaΣF₁=T-meg== 0(ii) Calculate the tension in the cable supporting the elevator.T= 17407.5Nfor 1.45 s.(iii) How high has the elevator moved during this time?Ay= 24.60X mm(c) The elevator decelerates at a rate of 0.65 for 2.9 s.s²(i) Newton's Law in the y-direction can be written as:Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards,pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration.ZF, T-mg= -1xmea

Given that: Mass of (elevator+person), mE=1650 kg

An elevator shown below filled with passengers has a mass of 1650 kg. The elevator does motions (a) through (c) in succession. T meg For each of the parts below draw a free body diagram of the elevator in your notebook for each of the parts (a) to (c). Draw the acceleration and velocity vectors in the boxes. For each part, are the vectors for tension in the string and weight of the elevator of equal lengths or unequal lengths.

An elevator shown below filled with passengers has a mass of 1650 kg. The elevator does motions (a) through (c) in succession. T meg For each of the parts below draw a free body diagram of the elevator in your notebook for each of the parts (a) to (c). Draw the acceleration and velocity vectors in the boxes. For each part, are the vectors for tension in the string and weight of the elevator of equal lengths or unequal lengths.

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter5: Newton's Law Of Motion

Section: Chapter Questions

Problem 72AP: Two small forces, F1=2.40i6.10tj N and F2=8.50i9.70j N , are exerted on a rogue asteroid by a pair...

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