Find the acceleration of the hanging block

[Emily_20]

2m₁2a₂/2-m2g=m₂a₂

 

[Emily_20]

I got a₂=m₂g/(2m₁-m₂)

 

[Satvik Pandey]

No.
I think I lead you on a wrong path. Sorry for that.:s
If m2 is pulled by 'x' distance then m1 is pulled by '2x' distance.

So $a_{1}=2a_{2}$.
Just use this value and find the answer.

 

[Emily_20]

2(m₁2a₂)-m₂g=-m₂a₂

 

[Satvik Pandey]

2(m₁2a₂)-m₂g=-m₂a₂

Correct. Go ahead.

 

[Emily_20]

4m₁a₂-m₂g=-m₂a₂

a₂=m₂g/(4m₁+m₂)

 

[Satvik Pandey]

4m₁a₂-m₂g=-m₂a₂

a₂=m₂g/(4m₁+m₂)

That looks good to me.:)

 

[Emily_20]

Thank you so much! you're my genius hero lol. Thank you everyone else who helped me too I appreciate it very much you saved me!

 

[Emily_20]

I understand everything so far but I am confused about this part:" If m2 is pulled by 'x' distance then m1 is pulled by '2x' distance.

So a1=2a2." how did you know that the distance is 2x?

 

[Emily_20]

a2=m2g/(4m1+m2) is the final answer correct?

 

[Satvik Pandey]

Thank you so much! you're my genius hero lol. Thank you everyone else who helped me too I appreciate it very much you saved me!

You are welcome.:D

 

[Satvik Pandey]

I understand everything so far but I am confused about this part:" If m2 is pulled by 'x' distance then m1 is pulled by '2x' distance.

So a1=2a2." how did you know that the distance is 2x?

This part is confusing too.

If m2 is lowered down by 'x' then string on the left side of the pulley is increased by 'x' and string on the right side of the pulley is also increased by 'x'.
Hence, m1 covers distance 2x if m2 is lowered down by distance 'x'.

There are two pulleys here I am talking about the movable pulley.

 

[Satvik Pandey]

a2=m2g/(4m1+m2) is the final answer correct?

Yes, $a_{2}=\frac{m_{2}g}{4m_{1}+m_{2}}$

 

[Emily_20]

Thank you, your awesome!

 

[Satvik Pandey]

Thank you, your awesome!

You are welcome.:)