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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)
[tex] \sqrt{ \frac{6d}{(g + a)} } [/tex]
2)
[tex]\sqrt{ \frac{6d}{g} }[/tex]
3)
[tex]\sqrt{ \frac{6d}{(g - a)} }[/tex]
4)
[tex] \frac{6d}{( g+ a)} [/tex]
5)
[tex]\sqrt{ \frac{2d}{(g + a)} }[/tex]
​

Please help A pulley system is hanged in a lift as shown below in thefigure A and B are two masses each having m and 2m massrespectively If d is the distance fr class=

Respuesta :

jcherry99

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

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