A row of milk cartons are lined up with masses of 0.280 kg, 0.550 kg and 1.08 kg. The milk cartons are then pushed forward by a 8.80 N force applied to the 0.280 kg milk carton. If the counter is frictionless, how much force does the 0.550 kg milk carton exert on the 0.280 kg milk carton?
Question
A row of milk cartons are lined up with masses of 0.280 kg, 0.550 kg and 1.08 kg. The milk cartons are then pushed forward by a 8.80 N force applied to the 0.280 kg milk carton. If the counter is frictionless, how much force does the 0.550 kg milk carton exert on the 0.280 kg milk carton?
Solution
To solve this problem, we need to use Newton's second law of motion, which states that the force exerted on an object is equal to its mass times its acceleration (F = ma).
Step 1: Calculate the total mass of the milk cartons. The total mass (m_total) is the sum of the masses of the three milk cartons, which is 0.280 kg + 0.550 kg + 1.08 kg = 1.91 kg.
Step 2: Calculate the total acceleration of the milk cartons. The total force (F_total) is 8.80 N, so the total acceleration (a_total) can be calculated by rearranging the formula F = ma to a = F/m. Therefore, a_total = F_total / m_total = 8.80 N / 1.91 kg = 4.61 m/s².
Step 3: Calculate the force that the 0.550 kg milk carton exerts on the 0.280 kg milk carton. The force that the 0.550 kg milk carton exerts on the 0.280 kg milk carton (F_550) is equal to the mass of the 0.550 kg milk carton times the total acceleration. Therefore, F_550 = m_550 * a_total = 0.550 kg * 4.61 m/s² = 2.54 N.
So, the 0.550 kg milk carton exerts a force of 2.54 N on the 0.280 kg milk carton.
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