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Block a rests on top of block b as shown in (figure 1). The table is frictionless but there is friction (a horizontal force) between blocks a and b. Block b has mass 6. 00 kg and block a has mass 2. 00 kg. If the horizontal pull applied to block b equals 12. 0 n, then block b has an acceleration of 1. 80 m/s2.

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The acceleration of block A is 0.6 m/s².

Net force on the blocks

The net force on the blocks is determined by applying Newton's second law of motion.

∑F = ma

Acceleration of bock A

The acceleration of block A is calculated by applying Newton's second law of motion as follows;

[tex]F_a + F_b = F\\\\F_a = F-F_b\\\\m_a a_a = F- m_ba_b\\\\a_a = \frac{F- m_ba_b}{m_a}[/tex]

Where;

  • Mb is mass of block B = 6 kg
  • Ma is mass of block A = 2 kg
  • [tex]a_b[/tex] is acceleration of block B = 1.8 m/s²
  • [tex]a_a[/tex] is the acceleration of block A = ?

[tex]a_a = \frac{12\ - \ 6 \times 1.8}{2} \\\\a_a = 0.6 \ m/s^2[/tex]

Thus, the acceleration of block A is 0.6 m/s².

Learn more about Newton's second law of motion here: https://brainly.com/question/3999427

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