For m2: t 1-m2g? sinα=m2a 1

So, m 1g? T 1T 1？ m2g？ sinα=m 1m2？ t 1 = m 1m2g+m 1m2g？ sinαm 1+m2

Second time: for m2: m2g-T2 = m2a2,

For m 1: t2-m 1g? sinα=m 1a2，

So, m2g? T2T2？ m 1g？ sinα=m2m 1？ T2=m 1m2g+m 1m2g？ sinαm 1+m2，

So: t1t2 =1;

(2) the first time: for m1:m1g-t1= m1a1.

For m2: t 1-m2g? sinα=m2a 1，

So, A 1 = M 1g? m2g？ sinαm 1+m2，

After increasing the mass of m2, for m2: m2 ′ g? sinα？ T3 = m2’a3，

For m1:t3-m1g = m1a3,

So, a3 = m2' g? sinα？ m 1gm 1+m2 '

According to: s = 12at2, we can get:

A has been praised, stepped on. What's your comment on this answer? Put away the comments