Finding tension and the time taken

[speedy46]

Homework Statement

A mass of 15 kg is supported at 2 meters from the ground by a light rope which passes over a light, smooth pulley and it carries at its over end a mass of 6kg which is held to the floor. Find the tension in the rope when the restraint is removed and the time taken for the 15kg mass to reach the floor.

Homework Equations

F net = T-m1g
F net = T+M2g

The Attempt at a Solution

F net = T-m1g
F net = T+m2g

-T+m2g=m2a

T=m1g+m1a

-(m1g+m1a)+m2g=M2a

a = (m2 - m1) ÷ (m1 + m2) × 9.8m/s^2

If m2 = 15 kg , m1= 6kg, g = 9.8m/s^2

a = (15kg - 6kg) ÷ (6kg + 15kg) × 9.8m/s^2

The acceleration of the both blocks is 4.2m/s^2

I would like to know how to work out the time taken for the 15kg block to hit the floor.

I would like to know if the following equation would work for the tension question.
Tension = m * g * sin + m * g

 

[tiny-tim]

Hi speedy46!

A mass of 15 kg is supported at 2 meters from the ground by a light rope which passes over a light, smooth pulley and it carries at its over end a mass of 6kg which is held to the floor. Find the tension in the rope when the restraint is removed and the time taken for the 15kg mass to reach the floor.
…
-T+m2g=m2a

T=m1g+m1a
…
a = (15kg - 6kg) ÷ (6kg + 15kg) × 9.8m/s^2

The acceleration of the both blocks is 4.2m/s^2

Yes, that's fine.

I would like to know how to work out the time taken for the 15kg block to hit the floor.

uhh? you have the acceleration … so just use one of the standard constant acceleration equations.

I would like to know if the following equation would work for the tension question.
Tension = m * g * sin + m * g

sin? sin of what?

what are you copying from?

just use the equations you wrote above …

-T+m2g=m2a

T=m1g+m1a