(B) The apparent weight $W'$ of a person in an elevator is given by $W' = m(g \pm a)$,where $m$ is the mass of the person,$g$ is the acceleration due

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The elevator moves downward with constant acceleration.

(B) The apparent weight $W'$ of a person in an elevator is given by $W' = m(g \pm a)$,where $m$ is the mass of the person,$g$ is the acceleration due to gravity,and $a$ is the acceleration of the elevator.When the elevator moves downward with an acceleration $a$,the apparent weight is $W' = m(g - a)$.Since $(g - a) Therefore,the person feels lighter when the elevator moves downward with constant acceleration.

Class 11 Physics Newton's Laws of Motion and Friction Apparent weight and Pseudo Force

Easy

$A$ person is standing in an elevator. In which of the following situations does he find his weight to be less than his actual weight?

AIIMS 2005 All AIIMS

A
The elevator moves upward with constant acceleration.
B
The elevator moves downward with constant acceleration.
C
The elevator moves upward with uniform velocity.
D
The elevator moves downward with uniform velocity.

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Newton's Laws of Motion and Friction

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Apparent weight and Pseudo Force

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$A$ person in an elevator accelerating upwards with an acceleration of $2 \, m/s^2$ tosses a coin vertically upwards with a speed of $20 \, m/s$. After how much time will the coin fall back into his hand (in $, s$)? $(g = 10 \, m/s^2)$

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$A$ person is standing in an elevator. In which of the following situations does he find his weight to be less than his actual weight?

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$A$ coin is dropped in a lift. It takes time $t_1$ to reach the floor when the lift is stationary. It takes time $t_2$ when the lift is moving up with constant acceleration. Then:

$A$ person of mass $50 \, kg$ stands on a weighing scale in a lift. If the lift is descending with a downward acceleration of $9 \, m/s^2$,what would be the reading of the weighing scale (in $, kg$)? (Take $g = 10 \, m/s^2$)

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$A$ mass $M$ is placed on a smooth wedge of angle $\theta$ resting on a frictionless surface. To keep the mass $M$ at rest relative to the wedge,the wedge must be given a horizontal acceleration $a$ equal to:

$A$ person in an elevator accelerating upwards with an acceleration of $2 \, m/s^2$ tosses a coin vertically upwards with a speed of $20 \, m/s$. After how much time will the coin fall back into his hand (in $, s$)? $(g = 10 \, m/s^2)$

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