Two children, A and B, are hanging on a massless, frictionless pulley as shown. Child A is standing on the ground, and child B is climbing up the rope. The mass of A is 50 kg, and the mass of B is 40 kg. Find the acceleration with which child B can climb the rope upwards such that child A does not rise from the ground.

1. Problem Description

Two children, A and B, are hanging on a massless, frictionless pulley as shown. Child A is standing on the ground, and child B is climbing up the rope. The mass of A is 50 kg, and the mass of B is 40 kg. Find the acceleration with which child B can climb the rope upwards such that child A does not rise from the ground.

2. Solution Steps

Let $m_A$ be the mass of child A and $m_B$ be the mass of child B.

$m_A = 50 \text{ kg}$

$m_B = 40 \text{ kg}$

Let $T$ be the tension in the rope. For child A not to rise from the ground, the tension in the rope must be equal to the weight of child A, so $T = m_A g$.

Now, consider the forces acting on child B. Let $a$ be the acceleration with which child B climbs up the rope. The forces acting on child B are the tension $T$ upwards and the weight $m_B g$ downwards.

The net force on child B is $T - m_B g$. According to Newton's second law, the net force equals mass times acceleration. Since child B is accelerating relative to the ground, we consider the acceleration $a$ relative to the ground. The equation for child B is:

$T - m_B g = m_B a$

Substituting $T = m_A g$, we get:

$m_A g - m_B g = m_B a$

$a = \frac{(m_A - m_B)g}{m_B}$

Using $g = 10 \text{ m/s}^2$:

$a = \frac{(50 - 40) \times 10}{40}$

$a = \frac{10 \times 10}{40}$

$a = \frac{100}{40}$

$a = 2.5 \text{ m/s}^2$

3. Final Answer

2.5 m/s^2