2.1 State Newton's Second Law of Motion in words - NSC Physical Sciences - Question 2 - 2024 - Paper 1

To create the free-body diagram for block A, consider the following forces:

1. The applied force $F$ at an angle of 50° to the horizontal.
2. The tension force $T$ acting upwards on the block due to block B.
3. The normal force $F_N$ acting upward on block A due to the surface.
4. The kinetic frictional force $F_f$ acting against the direction of motion (to the right).
5. The gravitational force $F_g$ acting downwards on block A, calculated as $F_g = m_A imes g = 4.1 ext{ kg} imes 9.8 ext{ m/s}^2$.

The forces must be labelled accordingly in the diagram.

2.3.1 Kinetic frictional force exerted on block A

The formula for kinetic frictional force is: $F_f = ext{µ}_k imes F_N$ where $ext{µ}_k = 0.35$ is the coefficient of kinetic friction. The normal force $F_N$ can be calculated as: $F_N = F_g - T ext{ (where T is the tension in the string)}$ Assume that $T$ can be obtained from block B's weight balancing with the forces acting on block A along with the applied force. After calculations, if $F_N$ is found to be X, then: $F_f = 0.35 imes X$

2.3.2 Acceleration of block A, by applying Newton's Second Law to each block separately.

Applying Newton's Second Law: For block A: $F_{net} = F - T - F_f = m_A imes a$ Expressing in magnitude: $49 imes ext{cos}(50^ ext{o}) - T - F_f = 4.1a$

For block B: $T - (2.3 ext{ kg} imes 9.8 ext{ m/s}^2) = 2.3 imes a_B$

Through simultaneous equations, you can solve for the accelerations for both blocks.