Zach, whose mass is 80 kg, is in an elevator descending at 10 m/s. The elevator takes 3.0 s to brake to a stop at the first floor. (a) What is Zach's weight before the elevator starts braking?

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Hi everyone in this practice problem, we will have to calculate the weight of two different patients before the elevator which the two patients is in begins to slow down. The two patients X and Y are 60 kg and 85 kg respectively. And they're both uh placed in an elevator descending at a rate of 15 m per second. Patient, X exits the elevator on the sixth floor. Suppose it takes the elevator eight seconds to stop on the sixth floor. We will have to calculate the weight of both patients before the elevator will begin to slow down. The options given are A X equals 588 Newton Y equals 739 Newton B X equals 452 Newton Y equals 833 Newton C X equals 588 Newton, Y equals 833 Newton. And lastly D X equals 585 Newton and Y equals to 833 Newton. So both patients will have the same free body diagram when we assume both of them to be acting like a particle. So in the vertical direction, both patients will had only experience the normal force F N and also the gravitational force F G, which will be equals to each patient's weight. So I'm gonna represent that with W X or Y just like so awesome. So since the elevator is not accelerating before it begins breaking, or it will remain at a constant rate before the break start, then in that case, A will equals to zero m per second squared or A Y. If you're looking only in the vertical direction will equals to zero, we're being asked to calculate the weight of both patients before the break. So A Y will equals to zero because of this. Then using Newton's second law, we can determine that Sigma F Y which will equals to M multiplied by A Y will equals to zero. Sigma F Y is going to be represented by F N minus F G taking F N or anything going upwards in the convention of positive will equals to zero. So then F N will equals to F G which will equals to the weight or W namely the normal force will be equal to the actual weight of persons, X or person Y. Thus, then we can calculate F and X will then equals to W X which will equals to M X multiplied by G. And that will give us M X is going to be 60 kg. G is going to be 9.81 m per second squared. And that will be F and X will then equals to 588 Newton. Next, similarly, we will have F N Y and that will equals to W Y, which will equals to M Y multiplied by G D M Y is given to be 85 kg and G is still going to be 9.8481 point meters per second squared. And that will give us F N Y to be Newton. So that will be the answer to this particular practice problem. So the weight of patient X before the elevator begins to slow down is going to be 588 Newton. And the weight of patient Y before the elevator begins to slow down is going to be 833 Newton, which will both correspond to option C in our answer choices. So answer C with X equals 588 Newton and Y equals 833 Newton will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.

Zach, whose mass is 80 kg, is in an elevator descending at 10 m/s. The elevator takes 3.0 s to brake to a stop at the first floor. (a) What is Zach's weight before the elevator starts braking?

Verified Solution

Key Concepts

Weight

Mass

Gravitational Acceleration