First slide

Different types of Forces

Question

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]

Moderate

A

5.89 N
B

57.1 N
C

67.1 N
D

None of these

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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