Question

person is standing on weighing machine placed on floor of elevator. The elevator starts going up with some acc, and then it moves with uniform vel for a while and finally decelerates to stop. The maximum and minimum weights recorded is 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the deceleration are the same, find true weight of the person and the magnitude of the acceleration.
Take g= 9.9 m/s^2.​

Answer 1

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person is standing on weighing machine placed on floor of elevator. The elevator starts going up with some acc, and then it moves with uniform vel for a while and finally decelerates to stop. The maximum and minimum weights recorded is 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the deceleration are the same, find true weight of the person and the magnitude of the acceleration.

Take g= 9.9 m/s^2.

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weighing machine shows maximum weight when elevator moves up with some acceleration.

⇒ m(g+a) = 72 g ➖➖➖➖➖➖[1]

weighing machine shows minimum weight when elevator moves down with some acceleration.

⇒ m(g-a) = 60 ➖➖➖➖➖➖➖➖[2]

∴ adding 1 & 2 , we get:-

⇒ 2 m g = 132 g

⇒ m= 66 kg

∴ weight = 66 kg

we know that:-

m (g-a) = 60 g

⇒ 66 ( g-a ) = 60 g

⇒ g-a = 60÷66 g

⇒ 11g-11a= 10g

⇒ g=11a

⇒ a = 9.9÷11=0.9

∴ accln = 0.9 m/ s²

Answer 2

Answer:

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Consider the acceleration of the elevator first upward and then downward in the same magnitude is a

And m be the mass of the person standing on the scale.

Then,

mg+ma=72g

And

mg−ma=60g

Now solving ,

M = $\frac{132g}{2g}$

$= 66kg$

$= 0.9ms {}^{ - 2}$