A 4 kg block on a horizontal, rough surface is connected to an 8 kg block by a light, inelastic string that passes over a frictionless pulley as shown below - NSC Physical Sciences - Question 6 - 2016 - Paper 1

Using Newton's second law, we can find the acceleration (a) of the system. The forces acting on the 8 kg block and the 4 kg block must be considered.

For the 8 kg block (falling down): $m_2 g - T = m_2 a$ Substituting values: $8g - T = 8a$ For the 4 kg block (being pulled horizontally): $T - f_f = m_1 a$ Where, $f_f = ext{friction} = ext{coefficient} imes N = ext{coefficient} imes m_1g$ Here, $N = m_1 g = 4 imes 9.8 = 39.2 ext{ N}$, and the friction becomes: $f_f = 0.6 imes 39.2 = 23.52 ext{ N}$ Now the equations are:

1. For the 8 kg block:
   $8g - T = 8a$
   $T = 8g - 8a$
2. For the 4 kg block: $T - 23.52 = 4a$ Substituting T from the first equation into the second: $(8g - 8a) - 23.52 = 4a$ $8g - 23.52 = 12a$ Using $g = 9.8 ext{ m/s}^2$: $8(9.8) - 23.52 = 12a$ $78.4 - 23.52 = 12a$ $54.88 = 12a$ a = rac{54.88}{12} = 4.57 ext{ m/s}^2

Using the calculated acceleration, substitute back into one of the equations to find the tension (T). From the equation for the 8 kg block: $T = 8g - 8a$ $T = 8(9.8) - 8(4.57)$ $T = 78.4 - 36.56$ $T = 41.84 ext{ N}$

The frictional force (f_f) can be calculated using the equation: $f_f = ext{coefficient} imes N$ Where:

• Coefficient of kinetic friction = 0.6
• Normal force (N) = m_1 g = 4 imes 9.8 = 39.2 ext{ N} Thus: $f_f = 0.6 imes 39.2 = 23.52 ext{ N}$