A bob is hanging over a pulley inside a car through, a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration 'a' vertically. The tension in the string is equal to -

Question

A bob is hanging over a pulley inside a car through, a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration 'a' vertically. The tension in the string is equal to - (A) m √g2 + a2 (B) m √g2 + a2 - ma (C) m √g2 + a2 + ma (D) m (g + a)

Tardigrade Admin · Accepted Answer

Applying Newton’s law along string T - m √ a 2+ g 2= ma T = ma + m √ a 2+ g 2

$m g^{2} + a^{2}$

$m g^{2} + a^{2} - ma$

$m g^{2} + a^{2} + ma$

$m (g + a)$

Applying Newton’s law along string
$T - m a^{2} + g^{2} = ma$
$T = ma + m a^{2} + g^{2}$

mg2+a2​

mg2+a2​−ma

mg2+a2​+ma

m(g+a)

Applying Newton’s law along string T−ma2+g2​=ma T=ma+ma2+g2​

$m g^{2} + a^{2}$

$m g^{2} + a^{2} - ma$

$m g^{2} + a^{2} + ma$

$m (g + a)$

Applying Newton’s law along string
$T - m a^{2} + g^{2} = ma$
$T = ma + m a^{2} + g^{2}$