Final answer:
The force exerted on the 9 kg block by the 4 kg block is 8.37 N, calculated by using Newton's second law to determine the system's acceleration and then applying that acceleration to the 9 kg block.
Step-by-step explanation:
The student's question revolves around the forces acting on blocks sliding on a frictionless horizontal surface under the influence of gravity. The goal is to find the magnitude of the force exerted on the 9 kg block by the 4 kg block when two other forces, Fl = 20 N (to the left) and Fr = 7 N (to the right), are applied to the system. Since the surface is frictionless, the only forces affecting horizontal acceleration are Fl and Fr.
To find the acceleration of the entire system we use Newton's second law (F = ma), combining the net force (Fl - Fr) with the total mass (1 kg + 9 kg + 4 kg = 14 kg). The acceleration (a) for the system is therefore calculated as (20 N - 7 N) / 14 kg = 13 N / 14 kg ≈ 0.93 m/s2.
Using this acceleration, the force that the 4 kg block exerts on the 9 kg block (F34) can be calculated using F = ma again, where m is the mass of the 9 kg block and a is the acceleration we just found. Hence, F34 = 9 kg * 0.93 m/s2 ≈ 8.37 N.