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===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 is
F′−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.