Block A of mass m is connected to block B of mass 7.5 kg by a light inextensible rope passing over a frictionless pulley - NSC Physical Sciences - Question 2 - 2023 - Paper 1

In the free-body diagram for block B, the following forces must be illustrated:

1. Weight (gravitational force) acting downwards:

   • $F_g = m_B g$, where $m_B = 7.5 kg$ and $g = 9.81 m·s⁻²$.

2. Tension (T) in the rope acting upwards.

The weight force can be calculated using: $F_g = 7.5 imes 9.81 = 73.575\,N$

The diagram should label these forces accordingly with arrows indicating the direction of each force.

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be expressed as: $F = ma$ where $F$ is the net force acting on the object, $m$ is its mass, and $a$ is the acceleration produced.

For block B: We know the gravitational force acting on block B is: $F_{gB} = 7.5 imes 9.81 = 73.575\,N$ Applying Newton's second law: $F_{net} = F_{gB} - T = m_B a$ Substituting what we know: $73.575 - T = 7.5 \times 3.87603$ Solving this equation allows us to find T.

For block A: The tension T is equal to the force acting on block A (which is its weight plus the net force acting). We can use the same approach, but the equation will look something like: $T = m_A imes a$ This ultimately allows us to calculate the mass $m_A$.

To find the maximum height ($h$) reached by block A, we can apply the energy principle. The energy conservation equation gives:

$E_{initial} = E_{final}$

Where:

• Kinetic energy at the max height = 0
• Potential energy at the max height is what we will calculate.

Using the equation: $mgh = \frac{1}{2}mv^2$

where $v = 3.41 m·s^{-1}$ and $g = 9.81 m·s^{-2}$, we can rearrange to find $h$:

$h = \frac{v^2}{2g} = \frac{(3.41)^2}{2 \times 9.81}$ $h = \frac{11.6281}{19.62} \approx 0.592\,m$

Thus, the total height above the ground reached by block A will be: $0.592 + 1.5 = 2.09 m.$