Sure, let's break this down step-by-step to understand how to calculate the apparent weight of the man in each situation. The apparent weight is essentially the normal force exerted by the lift floor on the man, and this varies based on the acceleration of the lift.
1. Given Data:
- Mass of the man [tex]\( m = 55 \)[/tex] kg
- Acceleration due to gravity [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
2. Case (a): Lift is rising with an acceleration of [tex]\( 1 \, \text{m/s}^2 \)[/tex]
When the lift is accelerating upwards, the apparent weight is calculated by adding the lift's acceleration to the gravitational acceleration.
[tex]\[
a_{\text{rising}} = 1 \, \text{m/s}^2
\][/tex]
The total effective acceleration becomes [tex]\( g + a_{\text{rising}} \)[/tex].
[tex]\[
\text{Apparent weight} = m \times (g + a_{\text{rising}})
\][/tex]
Plugging in the numbers:
[tex]\[
\text{Apparent weight} = 55 \, \text{kg} \times (9.8 \, \text{m/s}^2 + 1 \, \text{m/s}^2) = 55 \, \text{kg} \times 10.8 \, \text{m/s}^2 = 594 \, \text{N}
\][/tex]
So, the apparent weight when the lift is rising is 594 N.
3. Case (b): Lift is going down with an acceleration of [tex]\( 0.5 \, \text{m/s}^2 \)[/tex]
When the lift is accelerating downwards, the apparent weight is calculated by subtracting the lift's acceleration from the gravitational acceleration.
[tex]\[
a_{\text{down}} = 0.5 \, \text{m/s}^2
\][/tex]
The total effective acceleration becomes [tex]\( g - a_{\text{down}} \)[/tex].
[tex]\[
\text{Apparent weight} = m \times (g - a_{\text{down}})
\][/tex]
Plugging in the numbers:
[tex]\[
\text{Apparent weight} = 55 \, \text{kg} \times (9.8 \, \text{m/s}^2 - 0.5 \, \text{m/s}^2) = 55 \, \text{kg} \times 9.3 \, \text{m/s}^2 = 511.5 \, \text{N}
\][/tex]
So, the apparent weight when the lift is going down is 511.5 N.
4. Case (c): Lift is falling freely under the action of gravity
When the lift is in free fall, it is accelerating downwards at the rate of gravitational acceleration [tex]\( g \)[/tex].
In free fall, the man and the lift are both accelerating downwards at the same rate, hence there's no normal force acting on the man from the lift floor. Therefore, the apparent weight is:
[tex]\[
\text{Apparent weight} = 0 \, \text{N}
\][/tex]
So, when the lift is falling freely, the apparent weight of the man is 0 N.
