Question 1 of 5
A car of mass 1300 kg is stopped by a constant horizontal braking force of 6.2 kN. (a) Show that the deceleration of the car is about 5 m s^{-2}.
Why: Using Newton's Second Law, \( F = ma \). Here, braking force F = 6.2 kN = 6200 N (decelerating force). Mass m = 1300 kg. Acceleration a = \( \frac{F}{m} = \frac{6200}{1300} \approx 4.77 \, \text{m s}^{-2} \), which is about 5 m s^{-2}. The negative sign indicates deceleration.
Question 2 of 5
A 30 kg learner is pushing a solid fixed wall with action force \( \vec{A} \) of 24 N and the wall pushes him with a force \( \vec{R} \) as indicated. What is the value of reaction \( \vec{R} \) acting by the wall on the learner?
A
24 N
B
30 N
C
0 N
D
720 N
Why: By Newton's Third Law, for every action there is an equal and opposite reaction. The force applied by the learner on the wall (24 N) is equal in magnitude and opposite in direction to the force applied by the wall on the learner. Thus, \( R = 24 \, \text{N} \), opposite to \( A \). Option A matches this.
Question 3 of 5
A car with a mass of 1000.0 kg accelerates from 0 to 90.0 km/h in 10.0 s. (a) What is its acceleration?
Why: Convert 90 km/h to m/s: \( 90 \times \frac{1000}{3600} = 25 \, \text{m/s} \). Using \( v = u + at \), with u=0, v=25 m/s, t=10 s: \( a = \frac{v - u}{t} = \frac{25}{10} = 2.5 \, \text{m/s}^2 \). This uses kinematic equations derived from Newton's Second Law.
Question 4 of 5
Explain Newton's laws of motion with examples.
Why: The correctAnswer provides a complete model response meeting 5-mark criteria: introduction, detailed points with examples and formula, conclusion. Word count ~250.
Question 5 of 5
In the situation where a 0.2 kg block is on a rough surface with applied force causing acceleration, calculate the applied force F if g=10 m/s² and a=100 m/s².
Why: Using Newton's Second Law: Net force = ma, but total applied force F overcomes weight and provides acceleration. F = mg + ma = (0.2)(10) + (0.2)(100) = 2 + 20 = 22 N.
