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 ($F_{y}$) of the 60 N force can be calculated using the sine function:

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

First, we need to find the normal force ($F_{N}$):

$F_{N} = mg - F_{y} = (6 ext{ kg})(9.8 ext{ m/s}^2) - 30 ext{ N} = 58.8 ext{ N}$

Now, calculate the frictional force ($F_{f}$):

$F_{f} = \mu_k imes F_{N} = 0.13 imes 58.8 = 7.644 ext{ N}$

The horizontal component ($F_{x}$) of the 60 N force can be calculated using the cosine function:

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

Using Newton's second law, we can calculate the net force:

$F_{net} = F_{x} - F_{f} = 51.96 ext{ N} - 7.644 ext{ N} = 44.316 ext{ N}$

Now we apply $F = ma$ to find the acceleration:

$a = \frac{F_{net}}{m} = \frac{44.316}{6} = 7.386 ext{ m/s}^2$

Decrease. As the angle increases, the vertical component of the force will increase and thus the normal force will decrease, resulting in a reduction of the frictional force.

When object A exerts a force on object B, object B simultaneously exerts an equal and opposite force on object A.

The free-body diagram should include arrows indicating all forces:

• Normal force ($F_{N}$) acting upwards
• Gravitational force ($F_{g}$) acting downwards
• Tension force ($F_{T}$) from the car pulling the caravan
• Frictional force ($F_{f}$) acting in the opposite direction of motion.