Boxes \( A \) and \( B \) are in contact on a horizontal fictionless surface (Figure 1). Box \( A \) has mass 21.0 kg and box \( B \) has mass 7.0 kg . A horizontal force of 130 N is exerted on box \( A \) \( A \).

We start by noting that since the boxes are in contact and there’s no friction, they will accelerate together when the 130 N force is applied to Box A. Step 1. Find the Total Acceleration The total mass of the system is   m_total = m_A + m_B = 21.0 kg + 7.0 kg = 28.0 kg. Using Newton’s second law:   a = F_net / m_total = 130 N / 28.0 kg ≈ 4.64 m/s². So, both boxes accelerate to the right with an acceleration of about 4.64 m/s². Step 2. Determine the Contact Force Between the Boxes Now, let’s examine Box B. The only horizontal force acting directly on Box B is the contact force from Box A. We call this force F_contact. Since Box B has a mass of 7.0 kg and accelerates at 4.64 m/s², we apply Newton’s second law to Box B:   F_contact = m_B × a = 7.0 kg × 4.64 m/s² ≈ 32.5 N. Thus, Box A exerts a contact force of about 32.5 N on Box B. Step 3. (Optional) Verify for Box A For Box A, two forces act in the horizontal direction: the applied 130 N to the right and the contact force from Box B (which pushes back on A) to the left. According to Newton’s second law for Box A:   Net force on A = 130 N – F_contact. Setting this equal to m_A × a:   130 N – F_contact = 21.0 kg × 4.64 m/s² ≈ 97.4 N. Solving for F_contact gives:   F_contact = 130 N – 97.4 N ≈ 32.6 N, which is consistent with our earlier calculation (any slight differences are due to rounding). In summary: • The acceleration of both boxes is approximately 4.64 m/s². • The contact force between the boxes (the force that Box A exerts on Box B) is about 32.5 N.