Question

A student, standing on a scale in an elevator at rest, sees that his weight is 840 N. As the elevator rises, his weight increases to 1050 N, then returns to normal. When the elevator slows to a stop at the 10th floor, his weight drops to 588 N, then returns to normal. Draw force diagrams for the student during beginning and end of his elevator ride. Determine the acceleration at the beginning and end of the trip.

Answer 1

As per FBD while its accelerating upwards

we can say that

$F_n - mg = ma$

here normal force is given as

$F_n = 1050 N$

$W = 840 N$

now mass is given as

$m(9.8) = 840$

$m = 85.7 kg$

now we will have

$1050 - 840 = 85.7 \times a$

$a = 2.45 m/s^2$

Now while accelerating downwards we can say by FBD

$mg - F_n = ma$

again plug in all values

$840 - 588 = 85.7 \times a$

$a = 2.94 m/s^2$