Question

3.
A passenger of mass 72.2 kg is riding in an
elevator while standing on a platform scale. What
does the scale read when the elevator cab is (a)
descending with constant velocity (b) ascending
with acceleration 3.20 ms??​

Answer 1

Answer:

Given mass 72.5

Gravity acceleration, g = 9.8 m/${{s}^{2}}$ Scale reading = apparent weight = R = ?

(i) While descending with constant velocity, a =0

R = mg

R = 72.2 x9.8

=>R = 707.56 N

(ii) While ascending with a = 3.2 m/${{s}^{2}}$

R = m(g + a)

R = 72.2 (9.8 + 3.2) = 938.6 N

Answer 2

Answer:

Explanation:

Answer a)When elevator is descending with constant velocity than acceleration will be zero

Let's N be apparent weight of body

N-mg=ma

N-72.2(9.8)=0

N=707.56 Newton=72.2 kg

Answer b)Now elevator is moving with acceleration 3.2 m/s

N-mg=ma

N-72.2(9.8)=72.2(3.2)

N=938.6 Newton=95.77 kg