Applications of Newton’s Laws – Apparent Weight PDF

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Applications of Newton’s Laws – Apparent Weight...

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Applications of Newton’s Laws – Apparent Weight You are standing on a scale in an elevator on the 4th floor of the science building. As the elevator begins to descend to the first floor, you notice that the scale reads only 85% of your weight. What is the acceleration of the elevator during that period of time? Your true mass is 82 kg.

ΣFy = FN – w = - ma y

Acceleration is down, elevator is moving down and is speeding up

FN = 0.85 mg

w = mg

0.85 mg – mg = - may

rearrange to get a y

ay = (0.85 mg – mg) / (- m)

divide through by -m

ay = (- 0.85 g – (- g)) = - 0.85 g + g ay = (- 0.85 (9.80 m/s2) + (9.80 m/s2)) = - 8.33 + 9.8 ay = 1.47 m/s2

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You are standing on a scale in an elevator on the ground floor of the Empire State Building. As the elevator begins to ascend to the top floor, you notice that the scale reads 100 kg. Your true mass is 82 kg. What is the acceleration of the elevator?

ΣFy = FN – w = may

Acceleration is up, elevator is moving up and speeding up

FN = 100 kg * 9.80 m/s2 = 980 N FN – mg = may 980 N – (82 kg * 9.80 m/s2) = (82 kg) ay ay = (980 N – 804.4) / 82 kg ay = 2.15 m/s2

As the elevator in the previous question approaches the top floor, the elevator quickly decelerates at a rate of 3.0 m/s 2. Your true mass is 82 kg. What would the scale read when the elevator is slowing down?

ΣFy = FN – w = -may ay = 3.0 m/s2

Acceleration is down, elevator is moving up but slowing down

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FN – mg = may FN = mg + may FN = (82 kg * 9.80 m/s2) - (82 kg) (3.0 m/s2) FN = 803.6 – 246 = 557.6 N Scale would read FN /g = 557.6 N / 9.80 m/s2 = 56.9 kg...