# You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg. The elevator starts from rest and travels upward

Question

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg. The elevator starts from rest and travels upward with a speed that varies with time according to v(t) = (3.0 m/s2)t + (0.20 m/s3)t2. When t = 4.0 s, what is the reading on the bathroom scale?

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1 year 2021-09-05T01:00:16+00:00 1 Answers 2 views 0

= 922N

Explanation:

m = 64kg

v(t) = (3.0 m/s²)t + (0.20 m/s³)t².

t = 4.0s

differentiation of v(t) so the acceleration is given by

$$a(t) = \frac{ dv(t)}{dt} \\ = 3.0 + 0.4t\\ t = 4s\\a(4) = 3.0 + 0.4(4)\\= 4.6ms^-^2$$

using Newton’s second law

F (net) = ΣFg = ma

⇒R – w = ma

R = m(a + g)

= 64 (9.81 + 4.6)

= 922N