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 relative to car vertically. The tension in the string is equal to

• A. $m(g+a)$
• B.
  
  $m\sqrt{g^2+a^2}$
  
• C. $m\sqrt{g^2+a^2}+ma$
• D. $m\sqrt{g^2+a^2}-ma$

Answer 1

The correct option is C $m\sqrt{g^2+a^2}+ma$

Draw a free body diagram of given problem.

Calculate tension in the string.

Applying newton's law perpendicular to string Therefore,

$mg\sin \theta =ma\cos \theta$

$\tan \theta =\frac{a}{g}$

Now, applying newtons law along string,

$T=m\sqrt{g^2+a^2}+ma$

Final anwer (c)