A 6 kg object is pulled with a force of 60 N at an angle of 30° across a rough surface, as shown in the diagram below - NSC Technical Sciences - Question 2 - 2021 - Paper 1

The vertical component of the force can be calculated using trigonometric functions. It is given by:

$F_{y} = F imes ext{sin}( heta) = 60 ext{ N} imes ext{sin}(30°) = 30 ext{ N}$

First, calculate the normal force (

$F_{N} = mg - F_{y}$

). Given that the weight of the object is:

$mg = 6 ext{ kg} imes 9.8 ext{ m/s}^2 = 58.8 ext{ N}$

Thus,

$F_{N} = 58.8 ext{ N} - 30 ext{ N} = 28.8 ext{ N}$

Now, the frictional force can be calculated as follows:

$F_{f} = μ_k imes F_N = 0.13 imes 28.8 ext{ N} = 3.744 ext{ N}$

The horizontal component can be calculated using:

$F_{x} = F imes ext{cos}( heta) = 60 ext{ N} imes ext{cos}(30°) = 51.96 ext{ N}$

To calculate the acceleration, use Newton's second law:

$F_{net} = ma$

Where the net force is:

$F_{net} = F_{x} - F_{f} = 51.96 ext{ N} - 3.744 ext{ N} = 48.216 ext{ N}$

Thus,

a = rac{F_{net}}{m} = rac{48.216 ext{ N}}{6 ext{ kg}} = 8.036 ext{ m/s}^2

The friction will DECREASE. As the angle increases, the vertical component of the applied force increases, which reduces the normal force. Since friction is proportional to the normal force, an increase in the angle leads to a decrease in friction.

Newton's Third Law states that for every action, there is an equal and opposite reaction.

A free-body diagram should show the following forces acting on the caravan:

• Gravitational force acting downward (weight)
• Normal force acting upward
• Tension force from the car pulling the caravan horizontally
• Frictional force acting opposite to the direction of motion

This diagram must include arrows indicating the direction of each force, labelled accordingly.