A 50.0-kg person steps on a scale in an elevator. The scale reads 5000 N. What is the magnitude of the acceleration of the eleva

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2 years ago

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1 answer:

zaharov [31] · Answer 1 · 2023-06-17T05:29:30+03:00

The magnitude of the acceleration of the elevator is 90m/s^2

Due to Newton's Law ∑ Forces in direction of motion is equal to mass

multiplied by the acceleration

We have here two forces 5000 N in direction of motion and the weight of the person in opposite direction of motion

The weight of the person is his mass multiplied by the acceleration of gravity

So ,

W = mg , where m is the mass and g is the acceleration of gravity

m = 50 kg and g = 9.8 m/s²

Substitute these values in the rule above

W = 50 × 9.8 = 490 N

The scale reads 5000 N

F = 5000 N , W = 490 N , m = 50kg

F - W = ma

5000 - 490 = 50 a

4510 = 50 a

Divide both sides by 50

a = 90.2 m/s²

Hence the acceleration is 90m/s^2

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