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Three identical blocks connected by ideal (massless) strings are being pulled along a horizontal frictionless surface by a horizontal force F⃗ F → . (Figure 1)The magnitude of the tension in the string between blocks B and C is T T = 3.00 N N . Assume that each block has mass m m = 0.400 kg k g .

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Three identical blocks connected by ideal (massless) strings are being pulled along a horizontal frictionless surface by a horizontal force F⃗ F → .

The magnitude of the Force is 4.5 N

The objective of the question is to determine the magnitude of the force.

From the given information;

  • Since the tension in the string of blocks B and C is drawing blocks A and B, Then, acceleration will be equal.

∴

[tex]\mathbf{F=(m_A + m_B) a}[/tex]

where;

  • Force F = 3.00 N
  • mass = 0.400 kg

[tex]\mathbf{3=(0.400 + 0.400) a}[/tex]

[tex]\mathbf{a = \dfrac{3}{0.800 }}[/tex]

a = 3.75 m/s²

Now, the magnitude of the Force in the whole system is determined as follows:

  • Recall that:
  • F = ma

∴

[tex]\mathbf{F = (m_a+m_b+m_c )a}[/tex]

F = (0.400 +0.400+0.400) × 3.75

F = 1.2 × 3.75

F = 4.5 N

Therefore, we can conclude that the magnitude of the Force is 4.5 N

Learn more about Force here:

https://brainly.com/question/11697162?referrer=searchResults

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