Question

1. find apparent weight of a 280.0 lbs (pound) person when elevator goes up with acceleration of 1.60 m/s². (b) when elevator goes down with acceleration of 1.60 m/s². (c) when elevator goes with constant velocity up? find your answers by pound.(1lb=4.44newton)

Explanation:

Step1: Convert weight to mass

We know that weight \(W = mg\). Given \(W=280\space Lbs\), and \(1\space Lb = 4.44\space N\), so \(W=(280\times4.44)\space N\). Also, \(g = 9.8\space m/s^{2}\). Mass \(m=\frac{W}{g}=\frac{280\times4.44}{9.8}\space kg\).

Step2: Find apparent weight when accelerating up

The formula for apparent weight \(W_{a}\) when accelerating up is \(W_{a}=m(g + a)\). Substitute \(m=\frac{280\times4.44}{9.8}\space kg\), \(g = 9.8\space m/s^{2}\) and \(a = 1.60\space m/s^{2}\).

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First, \(\frac{280\times4.44}{9.8}\approx127.2\). Then \(W_{a}=127.2\times11.4 = 1450.08\space N\). Convert back to pounds: \(W_{a}=\frac{1450.08}{4.44}\approx326.6\space Lbs\)

Step3: Find apparent weight when accelerating down

The formula for apparent weight \(W_{a}\) when accelerating down is \(W_{a}=m(g - a)\). Substitute \(m=\frac{280\times4.44}{9.8}\space kg\), \(g = 9.8\space m/s^{2}\) and \(a = 1.60\space m/s^{2}\)

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Since \(\frac{280\times4.44}{9.8}\approx127.2\), then \(W_{a}=127.2\times8.2=1043.04\space N\). Convert back to pounds: \(W_{a}=\frac{1043.04}{4.44}\approx234.9\space Lbs\)

Step4: Find apparent weight when moving at constant velocity

When moving at constant velocity \(a = 0\). Using \(W_{a}=m(g + a)\), since \(a = 0\), \(W_{a}=mg\). We know \(mg = 280\space Lbs\) (because when \(a = 0\), there is no additional or subtractive force due to acceleration)

Answer:

(a) \(326.6\space Lbs\)
(b) \(234.9\space Lbs\)
(c) \(280\space Lbs\)