Answer:
During the deceleration of an ascending elevator, the normal force on the feet of a passenger is greater than her weight.
During the deceleration of a descending elevator, the normal force on the feet of a passenger is less than her weight.
Explanation:
Normal force on her feet is the upward force due to gravitational force acting downwards.
During the deceleration of an ascending elevator, the normal force on the feet of a passenger is greater than her weight. This is because the elevator is moving upwards although the acceleration is reducing.
During the deceleration of a descending elevator, the normal force on the feet of a passenger is less than her weight. This is because the elevator is moving downwards although the acceleration is reducing.
Answer:
Elevators decelerate to stop at a floor. During the deceleration of an ascending elevator, the normal force on the feet of a passenger is smaller than her weight. During the deceleration of a descending elevator, the normal force on the feet of a passenger is larger than her weight.
Explanation:
When the elevator is moving upward the normal force can be represented in the expression below;
= mg+ma
when deceleration, the acceleration is opposite =-ve N is express as;
[tex]F_{N}[/tex] = mg-ma
Therefore the normal force is smaller than the weight.
On the contrary, when travelling downwards the elevator normal force can be represented as;
= mg-ma
for the deceleration a is also opposite = -ve
therefore =mg+ma
Here the normal force is larger than her weight.
Therefore when the lady is ascending the elevator her weight is smaller to the normal force and when she is descending the elevator, her weight is bigger when compared to the normal force.
