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4. A 70.0 kg boy and a 45.0 kg girl use an elastic rope while engaged in a tug-of-war on an icy,
frictionless surface. If the acceleration of the girl toward the boy is 2.25 m/s², find the
acceleration of the boy toward the girl. (3M law : m2 a2 = - ml al)
Given:
ml =
m2 =
a2=
Unknown:
al=?
Equation:
m2 a2= - ml al

Respuesta :

MrRoyal

Answer:

[tex]a_1 = 1.446m/s^2[/tex]

Explanation:

Given

[tex]m_1 = 70.0kg[/tex] -- Mass of the boy

[tex]m_2 = 45.0kg[/tex] -- Mass of the girl

[tex]a_2 = 2.25m/s^2[/tex] -- Acceleration of the girl towards the boy

Required

Determine the acceleration of the boy towards the girl ([tex]a_1[/tex])

From the question, we understand that the surface is frictionless. This implies that the system is internal and the relationship between the given and required parameters is:

[tex]m_1 * a_1 = m_2 * a_2[/tex]

Substitute values for [tex]m_1, m_2[/tex] and [tex]a_2[/tex]

[tex]70.0 * a_1 = 45.0 * 2.25[/tex]

Make [tex]a_1[/tex] the subject

[tex]\frac{70.0 * a_1}{70.0} = \frac{45.0 * 2.25}{70.0}[/tex]

[tex]a_1 = \frac{45.0 * 2.25}{70.0}[/tex]

[tex]a_1 = \frac{101.25}{70.0}[/tex]

[tex]a_1 = 1.446m/s^2[/tex]

Hence, the acceleration of the boy towards the girl is 1.446m/s^2

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