Answer: 2.52 N
Explanation:
Since, Block B needs to be dragged at constant speed, it means net force on it would be zero. The tension T would act in the string connecting the two blocks. Frictional force would act on both the contact surfaces of block B.
Refer to Free body diagram:
F - T - f₁ - f₂ = 0
⇒F =- T + f₁ + f₂
\(f_2 =\mu_k (m_A+m_B) g\)
\(f_1=\mu_k m_A g\)
From block A,
\(T = \mu m_Ag\)
Thus,
\(F = \mu m_Ag+\mu_k m_A g+\mu_k (m_A+m_B) g\)
Inserting the values:
\(F = 3\times 0.30\times 1.40 N+0.30 \times 4.20 N = 2.52 N\)
