A 20 kg block is placed on a rough surface inclined at 30° to the horizontal - NSC Physical Sciences - Question 2 - 2021 - Paper 1

The free-body diagram should include:

• The weight force acting downwards (W = mg)
• The normal force acting perpendicular to the inclined surface.
• The kinetic friction force (fk) acting down the slope.
• The applied force (F) acting up the incline. It is essential to label each force with appropriate arrows showing their directions.

Using the equation of forces:

For the block moving up at constant velocity, the net force (F_net) is zero. Therefore:

$F - F_k - W imes ext{sin}( heta) = 0$

Where:

• $W = mg = 20 imes 9.8 = 196$ N
• $W imes ext{sin}(30^ ext{o}) = 196 imes 0.5 = 98$ N
• The frictional force, $F_k = 18$ N.

Substituting values into the equation:

$F - 18 - 98 = 0$

Solving for F gives:

$F = 116 ext{ N}$

The net force acting on the block as it moves from X to Y can be calculated as:

$F_{net} = - F_k - W imes ext{sin}( heta)$

Where:

• Frictional force = 18 N
• Weight component acting down the incline = 98 N.

Thus, the net force is:

$F_{net} = - (18 + 98) = - 116 ext{ N}$

Using the kinematic equation:

d = v_i imes t + rac{1}{2} a t^2

Since the block comes to a stop:

• Final velocity (v_f) = 0
• Initial velocity (v_i) = 2 m/s,
• Time (t) can be calculated based on constant deceleration.

From the forces, using

$F_{net} = ma$

The total acceleration is: a = -rac{F_{net}}{m} = -rac{116}{20} = -5.8 ext{ m/s}^2

With this, we can compute distance: d = rac{(v_i^2 - v_f^2)}{2a} = rac{(2^2 - 0^2)}{2 imes 5.8} = 0.34 ext{ m}