Question

Please help!

A pulley system is hanged in a lift as shown below in the
figure. A and B are two masses each having m and 2m mass
respectively. If d is the distance from the floor to B and the
lift starts to move with acceleration a, the time taken by B to reach the floor is,
1)
$\sqrt{ \frac{6d}{(g + a)} }$
2)
$\sqrt{ \frac{6d}{g} }$
3)
$\sqrt{ \frac{6d}{(g - a)} }$
4)
$\frac{6d}{( g+ a)}$
5)
$\sqrt{ \frac{2d}{(g + a)} }$
​

Answer 1

Answer:

B

Explanation:

Find the acceleration of the system

m2>m1

further m2 = 2m

and m1 = m

The acceleration is going to be the downward force created by m2 divided by the total mass (m2 + m2)

a = (m2*g - m1*g) / (m1 + m2)

a = (2m - m)*g/ (2m + m)

a = m * g/3m

a = 1/3 * g

Formula for time

Givens

distance := d

acceleration : = a  

vi = 0

Kinematic Formula

d = vi * t + 1/2 a t^2

Solution

d = 0 + 1/2 * 1/3 g  * t^2

6d = g * t^2

t^2 = 6d/g

t = √(6d/g)

Answer.

B