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

Asked by maham237 @ 02/12/2020 in Physics viewed by 265 People

Is the passenger accelerating? If so, what are the magnitude and direction of the acceleration?

Answered by maham237 @ 02/12/2020

Answer:

$0.45 m/s^2$ downward

Explanation:

There are two forces acting on the passenger on the elevator:

- The upward normal force, R

- The weight of the passenger, W

Therefore we can write Newton's second law as

$R-W=ma$

where a is the acceleration of the passenger and m its mass. Here we took upward as positive direction, so that R is positive.

The mass of the passenger can also be written as

$m=\frac{W}{g}$

where g = 9.8 m/s^2 is the acceleration of gravity.

So the equation becomes

$R-W=\frac{W}{g}a$

And solving for a, we find the acceleration:

$a=\frac{g}{W}(R-W)=\frac{9.8}{650}(620-650)=-0.45 m/s^2$

and the negative sign means the direction is downward.

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