5. Three point-like objects have the following masses and coordinates $$ (x, y) $$ : 1) $$ 2.5 \mathrm{~kg},(5.0 \mathrm{~m}, 6.0 \mathrm{~m}) $$ 2) $$ 3.5 \mathrm{~kg},(-2.0 \mathrm{~m}, 4.0 \mathrm{~m}) $$ 3) $$ 4.0 \mathrm{~kg}(x, 4.0 \mathrm{~m}) $$ If the $$ x $$-component of the center of mass of the three-object system is $$ x \mathrm{~m}=1.2 \mathrm{~m} $$, what is $$ x $$ ? A. 2.8 m B. 1.6 m C. -0.30 m D. 3.4 m E. 0.75 m

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Let the center of mass $$ x $$-coordinate be given by

$$
x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}.
$$

We know the masses and $$ x $$-coordinates:

- $$ m_1 = 2.5 \, \text{kg} $$ at $$ x_1 = 5.0 \, \text{m} $$
- $$ m_2 = 3.5 \, \text{kg} $$ at $$ x_2 = -2.0 \, \text{m} $$
- $$ m_3 = 4.0 \, \text{kg} $$ at $$ x_3 = x $$

The total mass is

$$
m_1 + m_2 + m_3 = 2.5 + 3.5 + 4.0 = 10.0 \, \text{kg}.
$$

The $$ x $$-coordinate of the center of mass is

$$
1.2 = \frac{2.5(5.0) + 3.5(-2.0) + 4.0(x)}{10.0}.
$$

Calculating the known products:

$$
2.5(5.0) = 12.5 \quad \text{and} \quad 3.5(-2.0) = -7.0.
$$

Thus, we have

$$
1.2 = \frac{12.5 - 7.0 + 4.0x}{10.0} = \frac{5.5 + 4.0x}{10.0}.
$$

Multiply both sides by $$ 10 $$:

$$
12 = 5.5 + 4.0x.
$$

Subtract $$ 5.5 $$ from both sides:

$$
4.0x = 6.5.
$$

Divide by $$ 4.0 $$:

$$
x = \frac{6.5}{4.0} = 1.625 \, \text{m}.
$$

Rounded to one decimal, $$ x \approx 1.6 \, \text{m} $$, which corresponds to option B.
$$ x = 1.6 \, \text{m} $$, so the answer is B.

Three point-like objects have the following masses and coordinates :

If the -component of the center of mass of the three-object system is , what is ?
A. 2.8 m
B. 1.6 m
C. -0.30 m
D. 3.4 m
E. 0.75 m

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Three point-like objects have the following masses and coordinates : If the -component of the center of mass of the three-object system is , what is ? A. 2.8 m B. 1.6 m C. -0.30 m D. 3.4 m E. 0.75 m