
The magnitude of acceleration of both the blocks will be same and is given by
a=4+14gsin60∘−gsin30∘
=5203−5
a=(43−1) ms−1
Applying Newton's second law for second block (along the direction parallel to the incline on which it is placed)
T−5=1×a
⇒T=5+43−1
=4(3+1)N
