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

In the free-body diagram for block A, the following forces should be represented:

• The applied force $F$ acting at an angle of 50° from the horizontal, with a horizontal component $F_x = F imes ext{cos}(50°)$ and a vertical component $F_y = F imes ext{sin}(50°)$.
• The weight of block A ($W_A = m_A imes g = 4.1 imes 9.8$), acting downward.
• The normal force ($F_N$) acting upward, which balances the weight.
• The force of kinetic friction ($F_f$) opposing the motion, calculated as $F_f = ext{friction coefficient} imes F_N$.

To find the kinetic frictional force, we first calculate the normal force on block A. The normal force can be determined considering the vertical forces acting on it:

$F_N = W_A - F_y = 4.1 imes 9.8 - (49 imes ext{sin}(50°))$

The frictional force is then given by:

$F_f = ext{friction coefficient} imes F_N = 0.35 imes F_N.$

Substituting the values yields the magnitude of the kinetic frictional force.

For block A, applying Newton's Second Law:

1. The net force is given as:

$F_{net} = F_x - F_f$

2. The equation becomes:

$F_{net} = m_A imes a_A$

where $m_A = 4.1 ext{ kg}$ and $F_f$ is calculated from the previous part.

3. Solve for the acceleration $a_A$ to determine its value.