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

To find the acceleration (a) of the system, we first define the net force acting on both blocks.

For the 4 kg block:

• In the horizontal direction, the net force is: $F = T - f$ Substituting friction gives: $f = \mu \cdot N = 0.6 \cdot (4 \text{ kg} \cdot 9.8 \text{ m/s}^2) = 23.52 \, \text{N}$

So, the equation becomes: $F = T - 23.52 \text{ N}$

For the 8 kg block (in vertical direction):

• The weight acting downwards is: $W = 8 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 78.4 \, \text{N}$
• The tension acts upwards. The net force for the 8 kg block: $F = 78.4 - T$

Equating the two blocks gives: $\sum F = (4 + 8) a$ Substituting in the tensions leads to: $T - 23.52 = 4a\ ext{ and }\ 78.4 - T = 8a$

Now, adding both equations results in: $T = 23.52 + 4a$

\text{(1)} \ 12a = 54.88 \to a = 4.57 \, \text{m/s}^2$$

Using the acceleration calculated, we can find Tension (T):

From the equation derived above: $T = 23.52 + 4a$ Substituting a: $T = 23.52 + 4(4.57) = 23.52 + 18.28 = 41.8 \, \text{N}$

Using the coefficient of kinetic friction and the normal force, the magnitude of the frictional force (f) is:

$f = \mu N$ Where \mu = 0.6 and N = 4 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 39.2 , \text{N}$$

Thus: $f = 0.6 \cdot 39.2 = 23.52 \, \text{N}$