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The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N. Is the passenger accelerating? If so, what are the magnitude and direction of the acceleration?

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skyluke89

Answer:

[tex]0.45 m/s^2[/tex] 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

[tex]R-W=ma[/tex]

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

[tex]m=\frac{W}{g}[/tex]

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

So the equation becomes

[tex]R-W=\frac{W}{g}a[/tex]

And solving for a, we find the acceleration:

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

and the negative sign means the direction is downward.

Acceleration is defined as the rate of change of velocity with respect to time. The Megnitude of the acceleration is 0.45 m/sec².While the direction is downward.

What is acceleration?

The rate of change of velocity with respect to time is known as acceleration. Its unit is m/sec².It is a time-based quantity.

The given data in the problem is;

R is the upward normal force,

W is the weight of the passenger

From Newton's second law;

[tex]\rm R-W = ma \\\\[/tex]

m is the mass of passenger can be written as;

[tex]\rm m= \frac{W}{g}[/tex]

[tex]\rm R-W = \frac{W}{g} a \\\\\rm a= \frac{g}{W} (R-W) \\\\ \rm a= \frac{9.81}{650} (620-650)\\\\ \rm a= -0.45 m/sec^2[/tex]

- ve shows the direction is downward.

Hence the megnitude of the acceleration is 0.45 m/sec².While the direction is downward.

To learn more about the acceleration refer to the link;

https://brainly.com/question/1992383

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