An elevator starts from rest with a constant upward acceleration. It moves 2 m in the first 0.6 s. A passenger in the elevator is holding a 3 kg package by a vertical string. The tension in the string during acceleration is (Take $g = 9.8 m / s^{2}$ )

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Solution

$S = u t + \frac{1}{2} a t^{2}$
$\Rightarrow 2 = 0 + \frac{1}{2} \times a \times 0.6 \times 0.6$
$a = \frac{4 \times 100}{6 \times 6} = \frac{100}{9} m / s^{2}$
$T - m g = m a$
$T = m (g + a)$
$= 3 \times (9.8 + \frac{100}{9}) = 62.7 N$

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An elevator starts from rest with a constant upward acceleration. It moves 2 m in the first 0.6 s. A passenger in the elevator is holding a 3 kg package by a vertical string. The tension in the string during acceleration is (Take g=9.8 m/s2)

S=ut+12at2 ⇒ 2=0+12×a×0.6×0.6 a=4×1006×6=1009m/s2 T−mg=ma T=m(g+a) =3×(9.8+1009)=62.7 N

An elevator starts from rest with a constant upward acceleration. It moves 2 m in the first 0.6 s. A passenger in the elevator is holding a 3 kg package by a vertical string. The tension in the string during acceleration is (Take $g = 9.8 m / s^{2}$ )

$S = u t + \frac{1}{2} a t^{2}$
$\Rightarrow 2 = 0 + \frac{1}{2} \times a \times 0.6 \times 0.6$
$a = \frac{4 \times 100}{6 \times 6} = \frac{100}{9} m / s^{2}$
$T - m g = m a$
$T = m (g + a)$
$= 3 \times (9.8 + \frac{100}{9}) = 62.7 N$