In the diagram below, a 4 kg block lying on a rough horizontal surface is connected to a 6 kg block by a light inextensible string passing over a light frictionless pulley - NSC Physical Sciences - Question 5 - 2017 - Paper 1

To create the free-body diagram for the 6 kg block:

1. Draw the block as a box.
2. Indicate the downward force due to gravity, labeled as $F_g = mg = 6 kg imes 9.8 m/s^2$.
3. Add the tension in the string (upwards), labeled as $T$.
4. Clearly indicate the forces acting on the block, with $F_g$ directed downwards and $T$ directed upwards.

To calculate the work done by the gravitational force:

Using the formula:
$W = F_d imes d imes ext{cos}( heta)$ Where $F_d = mg = 6 kg imes 9.8 m/s^2$, $d = 1.6 m$, and $heta = 0^{ ext{°}}$ (since the force acts in the direction of the displacement).

Calculating: $W = (6 imes 9.8) imes 1.6 imes ext{cos}(0^{ ext{°}})$.

The work done by the gravitational force is: $W = 6 imes 9.8 imes 1.6 = 94.08 ext{ J}$.

To find the speed of the 6 kg block, we apply the work-energy principle. The total energy lost by the 6 kg block will equal the work done against friction and the kinetic energy gained by it:

1. Calculate the gravitational potential energy lost: $ext{PE}_{ ext{initial}} = mgh = (6)(9.8)(1.6).$

2. Calculate the work done against kinetic friction: $W_{k} = ext{friction force} = ext{coefficient} imes ext{normal force} = 0.4 imes (4 imes 9.8) = 15.68 ext{ N}$,
   which acts over $1.6 m$, so: $W_{k} = 15.68 imes 1.6 = 25.088 ext{ J}.$

3. Use energy balance: ext{PE}_{ ext{initial}} - W_{k} = rac{1}{2}mv^2,
   where $m = 6 kg$ and $v = ?$.

Solving: (6)(9.8)(1.6) - 25.088 = rac{1}{2}(6)v^2.

After simplifying and solving for $v$, we find: $v ext{ (final speed)} = 3.71 ext{ m/s}$.