Fnet=ma
Resolve vector forces into component forces x and y with y perpendicular to surface and x paralell to surface.
mass 1 = 6kg
mass 2 = 5kg
g=9.8m/s2
Θ =45degrees
Forces for Mass 1
m1g=(6kg)(9.8m/s2)=58.8N
F1,x=(mg)sin θ = (58.8)sin 45 =41.58N since the object is falling the sign is negative thus,
F1,x = -41.5N This force is also equal to Tension T since it is pulling on mass 2 as they are both in freefall on a frictionless surface.
Forces for Mass 2
F2,x=(mg)sin θ = (49)sin 45 =34.65N since the object is falling the sign is negative thus,
F2,x= -34.65N
using 2nd Law: Fnet = Max where M is total mass(m1+m2) and ax is acceleration along the surface.
Fnet,x = Max
F1,x+F2,x =Max
-41.58-34.65N=(11)a
x
-76.23N=(11)ax
-76.23N
You can check that each mass has the same acceleration along the surface by applying F=ma to each mass:
F1,x / m1 =ax
F2,x / m2 = ax
as you would expect because both object are in freefall their component accelerations are equal to each other and to:
(g) Sin 45 = (9.8)Sin 45= 6.93m/s2.