Answer:
Step-by-step explanation:
3.6.1 The free body diagram for block A would look like:
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(↓)mg
│
▼
────────────
│ │
│ │
│ A │
│ │ T
│ │
────────────
│
▼
F
where mg is the gravitational force acting on block A, T is the tension in the cord and F is the net force acting on block A.
3.6.2 The net force acting on block A is given by:
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F - f_k = ma
where f_k is the force of kinetic friction, and a is the acceleration of block A. We know that F = 6 N and f_k = 0.025 * mg, where mg is the gravitational force on block A. Therefore, we can rewrite the equation as:
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6 - 0.025 * mg = 3a
Now, we can use Newton's Second Law to find the force acting on block B:
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T - mg = 4a
We can solve these two equations simultaneously to find the acceleration of block B:
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T - mg = 4a (1)
6 - 0.025 * mg = 3a (2)
Multiplying equation (2) by 4/0.025 and adding it to equation (1) gives:
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T - 4mg = 64a
Substituting T = mg + 4a from equation (1) gives:
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mg + 4a - 4mg = 64a
-3mg = 59a
a = -3mg/59 ≈ -0.152 m/s²
The magnitude of the acceleration of block B is therefore |-0.152| ≈ 0.152 m/s².
3.6.3 To find the magnitude of force F, we can use the equation:
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F - f_k = ma
We know that f_k = 0.025 * mg and a = -0.152 m/s² (since the blocks are moving to the left). Therefore, we can write:
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F - 0.025 * mg = -3.6
Solving for F gives:
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F = 0.025 * mg - 3.6
Substituting mg = 3g gives:
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F = 0.025 * 3g - 3.6 ≈ -0.19 N
Therefore, the magnitude of force F is approximately |-0.19| ≈ 0.19 N. Note that the negative sign indicates that the force is acting to the left.