A row of blocks are lined up with masses of 1.26 kg, 2.16 kg and 4.34 kg. The masses are then pushed forward by a 19.3 N force applied to the 1.26 kg block. If the table is frictionless, how much force does the 2.16 kg block exert on the 4.34 kg block? 1.74 N 5.37 N 0.868 N 10.8 N
Question
A row of blocks are lined up with masses of 1.26 kg, 2.16 kg and 4.34 kg. The masses are then pushed forward by a 19.3 N force applied to the 1.26 kg block. If the table is frictionless, how much force does the 2.16 kg block exert on the 4.34 kg block? 1.74 N 5.37 N 0.868 N 10.8 N
Solution
The force exerted on the entire system is 19.3 N. Since the table is frictionless, this force will be equally distributed among the blocks according to their masses.
The total mass of the system is 1.26 kg + 2.16 kg + 4.34 kg = 7.76 kg.
The acceleration of the system can be found using Newton's second law, F = ma. Rearranging for acceleration gives a = F/m.
So, the acceleration of the system is a = 19.3 N / 7.76 kg = 2.49 m/s².
The force that the 2.16 kg block exerts on the 4.34 kg block is the mass of the 4.34 kg block times the acceleration of the system.
So, the force is F = ma = 4.34 kg * 2.49 m/s² = 10.8 N.
Therefore, the 2.16 kg block exerts a force of 10.8 N on the 4.34 kg block.
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