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two blocks of masses 7 kg and 3 kg are initially resting on a frictionless floor. then force f and a force of 2n acts on the 7 kg and the 3 kg blocks as shown. if the 7 kg block exerts a force of 4n on the 3 kg block, find the magnitude of the force f.

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Final answer:

The magnitude of the force F applied to the 7 kg block, taking into account Newton's Third Law of Motion and Newton's second law for both blocks, is determined to be 18 N.

Step-by-step explanation:

To find the magnitude of the force F applied to the 7 kg block, we first need to consider Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction.

Given that the 7 kg block exerts a force of 4 N on the 3 kg block, the 3 kg block must exert an equal and opposite force of 4 N on the 7 kg block. The 3 kg block also has an additional 2 N force acting on it.

Since the surface is frictionless, the only horizontal forces acting on the 3 kg block are the 4 N force from the 7 kg block and the 2 N external force. These forces add up to give the 3 kg block a total force of 6 N. To find the acceleration (a) of the 3 kg block, we use Newton's second law (F = ma):

F = ma

6 N = 3 kg * a

a = 6 N / 3 kg

a = 2 m/s²

Because the blocks are initially in contact, they will have the same acceleration. Now, applying Newton's second law to the 7 kg block:

F - 4 N = 7 kg * 2 m/s²

F - 4 N = 14 N

F = 14 N + 4 N

F = 18 N

Therefore, the magnitude of the force F applied to the 7 kg block is 18 N.

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