Question

A bob is hanging over a pulley inside a car moving with constant acceleration \( a \) directed horizontally as shown. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration \( a \) 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 â

• A. \( \frac{m g}{a^2} \)
• B. \( \frac{m g}{a} \)
• C. \( \frac{m g}{a + a^2} \)
• D. \( \frac{m g}{a + a^2} \)

Hint:

For a bob in motion inside a car with both vertical and horizontal acceleration, the tension in the string is a result of both accelerations and can be calculated using Newton's second law.

Answer 1

The Correct Option is C