TS EAMCET 19-July-2022 Shift 2 Solved Paper

Force applied on block A,F=10N
Acceleration produced in block A,
a=20m∕ s2
According to Newton's second equation of motion,
F=mAa
mA=‌

F

a

=‌

10

20

=0.5‌kg
Again, block A is kept in contact with block B, where mB=1.5‌kg. If a1 be the common acceleration in the block system and R be reaction force on block A and B, then from free body diagram of block A and B, Equation of motion of block A isF′−R=mA⋅a1
⇒20−R=0.5a1....(i)
For block B,
R=mBa1
R=1.5a1......(ii)
From Eqs. (i) and (ii), we have
20−1.5a1=0.5a1⇒20=2a1
⇒‌‌a1=10m∕ s2
∴ From Eqs. (ii), R=1.5a1=1.5×10=15 Hence, force on block B will be 15N.