A passenger of mass $72.2 k g$ is riding in an elevator while standing on a platform scale. What does the scale read when the elevator cab is (a) descending with constant velocity (b) ascending with acceleration $3.20 {m s}^{- 2}$

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Solution

scale reading by elevator $= m (\to g - \to a)$
Ans a -
Since acceleration is zero, reading $= m g = 72.2 \times 9.8 = 707.56 N N$

Ans b-
Here, $\to a = - 3.2 m / s^{2}$
So. readings $= m (9.8 + 3.2) = 72.2 \times 13 = 938.6 N$

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A passenger of mass 72.2 kg is riding in an elevator while standing on a platform scale. What does the scale read when the elevator cab is (a) descending with constant velocity (b) ascending with acceleration 3.20 ms−2

scale reading by elevator =m(→g−→a)Ans a -Since acceleration is zero, reading =mg=72.2×9.8=707.56NNAns b-Here, →a=−3.2m/s2So. readings =m(9.8+3.2)=72.2×13=938.6N

A passenger of mass $72.2 k g$ is riding in an elevator while standing on a platform scale. What does the scale read when the elevator cab is (a) descending with constant velocity (b) ascending with acceleration $3.20 {m s}^{- 2}$

scale reading by elevator $= m (\to g - \to a)$
Ans a -
Since acceleration is zero, reading $= m g = 72.2 \times 9.8 = 707.56 N N$

Ans b-
Here, $\to a = - 3.2 m / s^{2}$
So. readings $= m (9.8 + 3.2) = 72.2 \times 13 = 938.6 N$