Concepts Of Physics MCQ Edition [Volume 1]
- Exam
- Concepts Of Physics MCQ Edition [Volume 1]
- Subject
- Physics
- Chapter
- Newton's Laws of Motion
Concepts Of Physics MCQ Edition [Volume 1] · Physics · Newton's Laws of Motion
A person stands on a weighing machine placed on the floor of an...
A person stands on a weighing machine placed on the floor of an elevator. The elevator starts moving upwards with a certain acceleration, proceeds with a uniform velocity for a while, and finally decelerates to a stop. The maximum and minimum weights registered are and . Given that the magnitudes of the acceleration and the deceleration are identical, determine the magnitude of the acceleration. Take .
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
Let the true mass of the person be .
When the elevator accelerates upwards with acceleration , the apparent weight is maximum.
W_\max = m(g + a)
When the elevator decelerates upwards with acceleration , the apparent weight is minimum.
W_\min = m(g - a)
Taking the ratio of the maximum and minimum apparent weights:
\dfrac{W_\max}{W_\min} = \dfrac{m(g + a)}{m(g - a)} = \dfrac{72}{60}
Substituting :
When the elevator accelerates upwards with acceleration , the apparent weight is maximum.
W_\max = m(g + a)
When the elevator decelerates upwards with acceleration , the apparent weight is minimum.
W_\min = m(g - a)
Taking the ratio of the maximum and minimum apparent weights:
\dfrac{W_\max}{W_\min} = \dfrac{m(g + a)}{m(g - a)} = \dfrac{72}{60}
Substituting :