Two carts are in contact on a horizontal frictionless surface. Cart 'A' has a mass of 3 kg, and cart 'B' a mass of 2 kg. (a) Find the force F on 'A' needed to give the two carts an acceleration of 0.8 m/sec2 . What is then the force exerted on 'B' by 'A'? (b) If the carts are given an equal acceleration in the opposite direction by pushing on B, find the force exerted on 'A' by 'B'. Explain why the force exerted on 'B' by 'A' in part (a) is not equal to the force exerted by 'A' on 'B' in part (b).

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To solve this problem, we will use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration (F = ma). We will also consider that the two carts are in contact and experience the same acceleration.

a) To find the force F on cart 'A' needed to give the two carts an acceleration of 0.8 m/sec², we will focus on the net force acting on the system.

The net force on the system is equal to the mass of the system (the sum of the masses of cart 'A' and cart 'B') multiplied by the acceleration of the system.

Mass of the system = mass of cart 'A' + mass of cart 'B'
m = 3 kg + 2 kg = 5 kg

Net force on the system = mass of the system × acceleration
F_net = m × a = 5 kg × 0.8 m/sec² = 4 N

Considering that the two carts are in contact, the force exerted on 'B' by 'A' must be equal in magnitude and opposite in direction to the force exerted on 'A' by 'B'. Therefore, the force exerted on 'B' by 'A' is also 4 N.

b) If the carts are given an equal acceleration in the opposite direction by pushing on B, the force exerted on 'A' by 'B' can be determined by considering the net force acting on cart 'A'.

The net force on cart 'A' is equal to the mass of cart 'A' multiplied by its acceleration.

Net force on cart 'A' = mass of cart 'A' × acceleration
F_A = m_A × a = 3 kg × (-0.8 m/sec²) = -2.4 N

Here, we observe that the force exerted on 'A' by 'B' is -2.4 N, which is in the opposite direction as in part (a).

To address why the force exerted on 'B' by 'A' in part (a) is not equal to the force exerted by 'A' on 'B' in part (b), we need to consider Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

In part (a), when 'A' exerts a force on 'B', 'B' exerts an equal and opposite force on 'A'. This is because the two carts are in contact, and their interaction forces are of the same magnitude but in opposite directions.

In part (b), when 'B' exerts a force on 'A', 'A' exerts an equal and opposite force on 'B'. This is again due to Newton's third law. However, the mass of 'B' is different from the mass of 'A', resulting in different magnitudes for the forces. Hence, the force exerted on 'B' by 'A' in part (a) is not equal to the force exerted by 'A' on 'B' in part (b).