Question

The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N.
1. what is the magnitude of the acceleration?
2. what is the direction of the acceleration

Answer 1

Explanation:

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Answer 2

Given:

The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N.

To FInd:

1. Magnitude of acceleration.

2. Direction of acceleration.

Solution:

In this problem, we've assumed the upward direction to be positive.

We've taken acceleration due to gravity, $g=10m/s^2$

The normal reaction by the lift, $N=620N$

Weight of the person, $W=650N$

So his mass, $m=\frac{W}{g}=65kg$

The net force on the person, $F=W-N=30N$ downwards.

So acceleration, $a=\frac{F}{m}=\frac{30}{65}=0.46m/s^2$ downwards.

Hence, the acceleration of the person is $0.46m/s^2$ downwards.