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 true weight of the person. Take .
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
Let be the true mass of the person and be the magnitude of acceleration and deceleration.
When the elevator accelerates upwards, the apparent weight is maximum. The reading on the weighing machine is given by:
m_\max = \dfrac{m(g + a)}{g} = m\left(1 + \dfrac{a}{g}\right) = 72
When the elevator decelerates upwards, the apparent weight is minimum. The reading on the weighing machine is given by:
m_\min = \dfrac{m(g - a)}{g} = m\left(1 - \dfrac{a}{g}\right) = 60
Adding the two equations, we get:
When the elevator accelerates upwards, the apparent weight is maximum. The reading on the weighing machine is given by:
m_\max = \dfrac{m(g + a)}{g} = m\left(1 + \dfrac{a}{g}\right) = 72
When the elevator decelerates upwards, the apparent weight is minimum. The reading on the weighing machine is given by:
m_\min = \dfrac{m(g - a)}{g} = m\left(1 - \dfrac{a}{g}\right) = 60
Adding the two equations, we get: