Block P, of unknown mass, is placed on a rough horizontal surface - NSC Physical Sciences - Question 2 - 2018 - Paper 1

To calculate the acceleration of the 3 kg block, we can use the equations of motion:

We know:

• Initial velocity, $v_i = 0$
• Displacement, $Δy = 0.5 ext{ m}$
• Time, $t = 3 ext{ s}$

Using the equation: $Δy = v_i t + \frac{1}{2} a t^2$ Substituting the known values: $0.5 = 0 + \frac{1}{2} a (3)^2$ Solving for acceleration ($a$): $0.5 = \frac{1}{2} a (9)$ $a = \frac{0.5 \times 2}{9} = \approx 0.111 \text{ m/s}^2$

To find the tension ($T$) in the string, consider the forces acting on the 3 kg block:

• Weight of the block: $W = m g = 3 imes 9.8 = 29.4 ext{ N}$
• The net force acting on the block is given by Newton's second law: $F_{net} = m a = 3 imes 0.111 = 0.333 ext{ N}$

Now, according to the force balance: $W - T = F_{net}$ Substituting the values: $29.4 - T = 0.333$ Thus, $T = 29.4 - 0.333 = 29.067 ext{ N}$

Block P's free-body diagram should show:

• The weight ($W_P$) acting downward.
• The tension ($T$) acting horizontally due to the string. Label these forces clearly with their directions.

From the tension we found earlier ($T$ = 29.067 N) and the acceleration, we can calculate the mass of block P using: $T = m_P g$ Where:

• $g = 9.8 ext{ m/s}^2$ Thus, $m_P = \frac{T}{g} = \frac{29.067}{9.8} \approx 2.96 ext{ kg}$