# Tension at varying velocites

#### che

Hey need some help with the methodology on how to solve this question

a elevator has a mass of 2000kg is being lowered down a shaft. it movesfrom rest with an acceleration of 4m/s^2 until it is traveling at 15m/s. it then travels at a constant speed for 700m and comes to rest in 6s.Calculate the tension in the cable supporting the elevator during:
a) the initial period of acceleration
b)The period of constant speed travel
c) the final retardation period.

thanks guys.

#### werehk

PHF Hall of Fame
First, I want to ask you a couple of questions to see how much you know. If you draw a free-body diagram, what are the forces present? One you should know, that the tension.

What is the difference between travelling with constant speed and accelerating, is that related to Newton's Laws?

arbolis

#### che

Yeah when i was typing i started to piece together the solution ma=t-mg therefore T= m(g-a). constant speed = tension

Thanks for the reply anywho good luck with the AL exams

#### werehk

PHF Hall of Fame
Thanks for your best wishes. It's good that you can write down the related equations. So you can find that constant speed means the acceleration and so the weight would be the same as tension.

When the elevator moves down with an acceleration, it implies that the weight would be greater than the tension. So you can put the value inside. mg-T=ma

For c),retardation actually means that the acceleration is in opposition to the original motion. So we can see that in this case tension is greater than that of weight. So T-mg= ma in this case