It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor. What is the passenger's weight (a) Before the elevator starts moving?

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Hi everyone in this practice problem, we will have a newly designed elevator which will reach m per second in velocity just within eight seconds. And we're being asked to determine the weight of an 85 kg load before the elevator will begin to move. The options given are a 648 Newton B 245 Newton C 833 Newton and D 368 Newton. So in this case, the 85 kg load will be acted or will be uh assumed as a particle. And the free body diagram of that, which will be represented by this uh pictorial representation will only contain two different forces, which is going to be first, the normal force pointing upwards which is will be represented by F N and second, the gravitational force or the weight, which will be represented by F G. We have to determine the weight before this, the elevator starts moving, which will mean that the elevator will not be accelerating or in this case A of the elevator or A Y because we're looking at only the vertical direction here A Y will equals to zero m per second squared. And because of this then applying new 12th law, we will get that the Sigma or the net force acting on the Y axis will equals to M multiplied by A Y which will equals to zero because A Y equals to zero. So Sigma F Y, according to our free body diagram will only have F N and F G. And in this case, I am going to uh take the convention of anything going upwards as positive. So we have Sigma F Y equals F N minus F G which will equals to zero. So in this case, our F N will equals to F G just like so which will equals two W or weight, namely, the normal force will be equal to the actual weight of the load when the elevator is not moving. Thus, W will equals to M multiplied by G which will equals to 85 kg multiplied by 9.81 m per second squared, which will actually equals to 833 Newton. And in this case, as we have um achieved from the Newton second law, we have F N two B equals to W which will equals to 833 Newton, which is going to be the weight of the 80 85 kg load before the elevator began moving. So F N equals to 833 Newton will correspond to option C which will mean that option C is going to be the answer to this particular practice problem. So that'll 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.

It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor. What is the passenger's weight (a) Before the elevator starts moving?

Verified Solution

Key Concepts

Weight

Mass

Gravitational Acceleration