The engine of a car provides a forward force of 1240 N and the total resistive force on the car is 800N. If…

Analyze the question: A car starts from rest. A forward force of 1240 N acts on it while a resistive force of 800 N opposes the motion. We need the distance travelled before its speed reaches 4 m/s.

Concept: Newton's second law gives the acceleration from the net force, F_net = ma. Then we use the equation of motion v² = u² + 2as.

Solution:

1. Net force = 1240 − 800 = 440 N

2. Acceleration: a = F_net / m = 440 / 1220 ≈ 0.3607 m/s²

3. Use v² = u² + 2as with u = 0 and v = 4: 16 = 2(0.3607)s s = 16 / 0.7213 ≈ 22.18 m

Alternatively, using the work-energy theorem: Net force × distance = ½mv² 440 × s = ½ × 1220 × 16 = 9760 s = 9760 / 440 ≈ 22.2 m

Answer: The distance is about 22.2 m, which is option B.

Tip: In "net force" problems, always subtract the resistive force first. Then you can use either F = ma with the equations of motion, or the work-energy method (Fs = ½mv²). The work-energy method skips finding acceleration, so it saves time in the exam.