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Q.

A lift starting from rest with a constant upward acceleration moves 1.5 m in 004 s. If a person standing
in the lift holds a packet of 2 kg by a string, then the tension in the string due to motion is
[UP CPMT 2013]

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a

57.1 N

b

67.1 N

c

5.89 N

d

None of these

answer is B.

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Detailed Solution

Given, u = 0, s = 1.5 m, t = 0.4 s

From second equation of motion, $s = u t + \frac{1}{2} a t^{2}$

$\Rightarrow 1.5 = 0 + \frac{1}{2} a (0.4)^{2}$

$\Rightarrow a = \frac{1.5 \times 2}{(0.4)^{2}} = 18.75 {ms}^{- 2}$

As the string is moving upwards with this acceleration

$∴ T = m (g + a) = 2 (9.8 + 18.75) = 57.1 N$

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A lift starting from rest with a constant upward acceleration moves 1.5 m in 004 s. If a person standingin the lift holds a packet of 2 kg by a string, then the tension in the string due to motion is [UP CPMT 2013]