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ssm A 95.0 -kg person stands on a scale in an elevator. What is the apparent weight when the elevator is $\quad$ (a) accelerating upward with an acceleration of $1.80 \mathrm{m} / \mathrm{s}^{2}, \quad$ (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of 1.30 $\mathrm{m} / \mathrm{s}^{2} ?$

$1.10 \times 10^{3} \mathrm{N}$

931 $\mathrm{N}$

808 $\mathrm{N}$

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Sheffield

University of Winnipeg

To solve this question, we have to apply Newton's second law on disk over what is the normal force, because what what the scale can measure is the normal force. So the normal force is what the question is calling the apparent weight. Then I will choose this reference frame to serve both a B and C. Then it goes as follows. Newton's second law, thousands that the net force on the Y axis that is acting on the person is there close to the mass. Off the person times the acceleration off the person in this situation, the 1st 1 The acceleration is pointing your ports okay. And then the net force in the Y direction is composed by the normal force and the weight force. Remember, I'm applying. Utah cycled along to the person so the tension forced doesn't count because it acts on the elevator. So it goes normal force minus wait. Forced. Is it close to the mask off the person times its acceleration So the normal force is given by the weight 40 plus m times eight. Remember, that's the weight force is given by M times G show that the normal force is given by M times G plus m times A. So we can factor out em to get em times g plus a. So in this first item remembering that tree is approximately 9.8 meters per second squared. We got a normal force given by 95 times 9.8 plus 1.8, which results in approximately 1000 and 100. Or we can write it as 1.1 time. Stand to the third mutants. This is dance or for the first item. Now, for the second item, we have a constant velocity. Constant velocity implies zero acceleration. Then you don't. Second law tells us that the net forest in the Y direction is equals to the mass times acceleration, but acceleration are zero. So we have that the normal forest minus the weak force. Easy questions here and the normal force in the situation. Is it close to the weight force which is given by M times G 7 95 times 9.8 resulting in ah normal force off 931 mutants. This is the answer for the second item. Now, for the last item, we have a downward acceleration off 1.3 meters per second squared. So the second lot tells us that the net force in the Y direction is a close to minus the mass times acceleration because now that celebration is pointing downwards to the negative direction off my reference frame. So the normal force mine is the weight force is a close to minus the mass times acceleration. Then the normal force is given by the weight force minus the mass times acceleration. Remembering that the weight force is given by M times G, we got a normal force off M times G minus M times A and finally factoring out M, we got em times G minus eight and these results in 95 times 9.8 minus 1.3, which is approximately 808 Newtons. These is their last answer of this question.

Brazilian Center for Research in Physics