Back to 6.1 Solving Problems with Newton’s Laws

6.1 Exercise 20

Avg difficulty 2.5

An elevator filled with passengers has a mass of 1.70×103kg1.70 \times 10^{3} \text{kg}. (a) The elevator accelerates upward from rest at a rate of 1.20m/s21.20 \text{m}/\text{s}^{2} for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator accelerates opposite to the motion at a rate of 0.600m/s20.600 \text{m}/\text{s}^{2} for 3.00 s. What is the tension in the cable during acceleration opposite to the motion? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

OpenStax University Physics Volume 1, 6.1 Solving Problems with Newton’s Laws, exercise 20

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Adapted from University Physics Volume 1 © 2016 Rice University. Licensed under CC BY 4.0. Source: University Physics Volume 1. Imported at revision db44eb9.