Question

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,

Answer 1

Answer:

$a_1 = 1.446m/s^2$

Explanation:

Given

$m_1 = 70.0kg$ -- Mass of the boy

$m_2 = 45.0kg$ -- Mass of the girl

$a_2 = 2.25m/s^2$ -- Acceleration of the girl towards the boy

Required

Determine the acceleration of the boy towards the girl ($a_1$)

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:

$m_1 * a_1 = m_2 * a_2$

Substitute values for $m_1, m_2$ and $a_2$

$70.0 * a_1 = 45.0 * 2.25$

Make $a_1$ the subject

$\frac{70.0 * a_1}{70.0} = \frac{45.0 * 2.25}{70.0}$

$a_1 = \frac{45.0 * 2.25}{70.0}$

$a_1 = \frac{101.25}{70.0}$

$a_1 = 1.446m/s^2$

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