Q.8
A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. The variation of the height (in m) of the elevator, from the ground, with time (in s) is given by:
Taking the acceleration due to gravity , the maximum variation of the object’s weight (in N) as observed in the experiment is ______.
Answer: 2
Solution:
1. Understanding the Problem
An elevator moves vertically with its height described by a sinusoidal function:
The object has a mass , and .
We need to find the maximum variation in the object's apparent weight as observed on a scale inside the elevator.
2. Key Physics Concepts
Apparent Weight: The normal force exerted by the scale, which equals the object's apparent weight.
When the elevator accelerates upward, .
When the elevator accelerates downward, .
The variation in weight is due to the elevator's acceleration , derived from the second time derivative of .
3. Derive the Elevator's Acceleration
Given:
First Derivative (Velocity):
Second Derivative (Acceleration):
Substitute :
4. Determine Apparent Weight
The apparent weight varies with the acceleration :
The maximum variation occurs when reaches its extreme values ():
Maximum Weight:
Minimum Weight:
Variation in Weight:
5. Final Answer
The maximum variation of the object’s weight is:
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