A 360 kg crate rests on the back of a truck with a rough surface - NSC Technical Sciences - Question 2 - 2022 - Paper 1

Newton's First Law of Motion states that an object will continue in a state of rest or uniform velocity unless acted upon by a non-zero resultant force.

The free-body diagram includes the following forces:

1. Weight (W) acting downward: labeled as W = mg.
2. Normal force (N) acting upward: equal in magnitude and opposite in direction to the weight.
3. Frictional force (F_f) acting to the left: this opposes the motion caused by the braking.

The acceleration can be calculated using the formula: $a = \frac{\Delta v}{\Delta t}$ Where:

• Initial velocity, $v_i = 105 \text{ km/h} = 29.17 \text{ m/s}$
• Final velocity, $v_f = 62 \text{ km/h} = 17.22 \text{ m/s}$
• Time interval, $\\Delta t = 7 \text{ s}$ Thus, $a = \frac{17.22 - 29.17}{7} = -1.7 \text{ m/s}^2.$

Using Newton's second law $F_{net} = ma$, we can calculate the net force acting on the truck using: $F_{net} = (m_{truck} + m_{crate}) \cdot a$ Where:

• Total mass, $m_{total} = 4,550 \text{ kg} + 360 \text{ kg} = 4,910 \text{ kg}$ Substituting the values: $F_{net} = 4,910 \cdot (-1.7) = -8,396.7 \text{ N}$ Thus, the force exerted by the brakes is approximately $8,396.7 \text{ N}$ in the opposite direction of the motion.