A 518- physics student stands on a bathroom scale in an 846- (including the stud

Question

A 518- physics student stands on a bathroom scale in an 846- (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 464 .

(a)Find the magnitude of the acceleration of the elevator.
(b)Find the direction of the acceleration of the elevator.(upwards or downwards )
(C)What is the acceleration if the scale reads 677 ?
(D)If the scale reads zero, should the student worry? Explain.
Essay answers are limited to about 500 words (3800 characters maximum, including spaces).
(E)What is the tension in the cable in part A?
(F)What is the tension in the cable in part D?

Explanation / Answer

F = m a where F = Fs - m g where F = net force on student and Fs = force of scales on student a) a = (Fs - m g) / (W / g) since m = W/g (mass and weight of student) a = (464 - 518) g / 518 = -.10 g = -.98 m/s^2 b) since the acceleration of the student is downwards the acceleration of the elevator is downward c) a = (677 - 518) g / 518 = .31 g = 3.0 m/s^2 and the acceleration is upwards d) Fs = 0 then F = - m g and the student is in free fall since the only force on the student is the gravitational force e) (M + m) a = T - (M + m) g where T is tension and M g + m g is the total weight of student and elevator T = (M + m) g + (M + m) a T = 518 + 864 + (518 + 864) / 9.8 * -.98 = 1234 N f) substitute the acceleration as in (e) and solve you should get zero since we have the positive direction of a as upward and a = -g

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