Calculate the net acceleration produced in the arrangement shown below.

Explanation

(a) Net pulling force =3 mg-mg=2mg
Total mass =3m+m=4m
`a=(2mg)/(4m)=(g)/(2) `
(b) Pulling force= 2mg
Total mass =3m
`a=(2 mg)/(3m)=(2g)/(3)`
(c ) Net pulling force= 3mg-2mg=mg
Total mass =3m+m+2m=6 m
`a=(mg)/(6mg)=(g)/(6)`
(d) Net pulling force= 3 mg +mg-2 mg= 2 mg
Total mass =3m+m+2m=6m
`a=(2g)/(6mg)=(g)/(5)`
(e) Net pulling force = 4 mg `sin 30^(@)-mg`=mg
Total mass =4m+m= 5m
`implies a=(mg)/(5m)=(g)/(5)`