A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. Suppose that the variation of the height y (in m ) of the elevator, from the ground, with time t (in s

Question

A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. Suppose that the variation of the height $y$ (in m ) of the elevator, from the ground, with time $t$ (in s ) is given by $y = 4 [1 + sin (\frac{2 π t}{T})]$ , where $T = 40 π s$ . Taking acceleration due to gravity, $g = 10 m / s^{2}$ , the maximum variation of the object's weight (in N ) as observed in the experiment is $____$ .

Answer:

1

Solution:

$y = 4 [1 + sin (\frac{2 π t}{T})]$

$a_{ma x} = 4 (\frac{2 π}{T})^{2}$

$T = 40 π \Rightarrow a_{ma x} = 0.01 m / s^{2}$

$Δ W_{ma x} = 2 m a_{ma x} = 2 (50) (0.01) = 1 N$

TestHub · Accepted Answer

y = 4 [ 1 + sin ( T 2 π t ​ ) ] a ma x ​ = 4 ( T 2 π ​ ) 2 T = 40 π ⇒ a ma x ​ = 0.01 m / s 2 Δ W ma x ​ = 2 m a ma x ​ = 2 ( 50 ) ( 0.01 ) = 1 N

Physics - NLM Question with Solution | TestHub

Doubts & Discussion