# Question #99c13

May 4, 2016

#### Explanation:

Apologies for the rubbish diagram, but I find it usually helps:

Consider each in turn:
a) going up and speeding down. The person is going upwards, but his speed is decelerating. By Newton's 2nd law, this means there is a net resultant force in the opposite direction to that of travel i.e. a net force downwards, causing acceleration in a downwards direction (upwards deceleration). This causes the lift to slow down . Hence: $m g$ (weight) $> F$ (force on person from lift floor)

b) going down and slowing down. Again by Newton's 2nd law, this means there is a net resultant force in the opposite direction to that of travel (i.e. a net force upwards this time to cause the acceleration in an upwards direction to slow down the lift ). Hence $F > m g$

c) going up and speeding up. By Newton's 2nd law, this means there is a net force in the same direction to that of travel (i.e. a net force upwards this time to cause the acceleration upwards ). Hence $F > m g$

So in both (b) and (c) , the force on the person's foot exceeds their weight. This is answer (d)

PS is the wording of the question actually "feet" rather than foot which would make more sense and prevent ambiguity . Technically the force F is shared between the person's 2 feet (unless they only have one leg). So in the descriptions, F is the combined force in both feet. Oh, the joy of Physics!