A car weighs 1000 kg. It is moving with a uniform velocity of 72 km/h towards a straight road. The driver suddenly presses the brakes. The car stops in 0.2 s. The retarding force applied on the car to stop it is
- (a)100 N
- (b)1000 N
- (c)10 kN
- (d)100 kN
Correct — D, 100 kN. Convert the speed: 72 km/h = 72 x 1000/3600 = 20 m/s. The car goes from 20 m/s to rest in 0.2 s, so the deceleration is a = change in velocity / time = 20 / 0.2 = 100 m/s^2. By Newton's second law the retarding force is F = m x a = 1000 kg x 100 m/s^2 = 100000 N = 100 kN.
- (a)100 N — Too small by a factor of 1000 — it corresponds to a = 0.1 m/s^2, i.e. stopping over about 200 s, not 0.2 s.
- (b)1000 N — Corresponds to a = 1 m/s^2; that would take 20 s to stop the car, a hundred times the given 0.2 s.
- (c)10 kN — Corresponds to a = 10 m/s^2, which would need a stopping time of 2 s — ten times too long.
This is Newton's second law (F = m x a) combined with a bit of kinematics. First convert the speed from km/h to m/s, then find the deceleration from the change in velocity over the stopping time, and multiply by the mass. Equivalently, F = change in momentum / time.
Two steps hide the traps: forget to convert 72 km/h to 20 m/s and the force comes out wrong; and the answer (100 kN) feels surprisingly large because stopping in just 0.2 s means a very high deceleration of 100 m/s^2.
- 1 km/h = 1000 m / 3600 s; divide km/h by 3.6 to get m/s, so 72 km/h = 20 m/s.
- Deceleration a = change in velocity / time = 20 / 0.2 = 100 m/s^2.
- Newton's second law: F = m x a, equivalently F = m x (change in velocity) / time (rate of change of momentum).
- F = 1000 x 100 = 100000 N = 100 kN (1 kN = 1000 N).
- Skipping the 72 km/h to 20 m/s conversion.
- Mixing up newtons and kilonewtons (1 kN = 1000 N).
Asked as a numeric F = ma / impulse problem with a km/h to m/s conversion built in.
A car is running on a road at a uniform speed of 60 km/hr. The net resultant force on the car is
- (a) Driving force in the direction of car's motion
- (b) Resistance force opposite to the direction of car's motion
- (c) An inclined force
- (d) Equal to zero
Answer(d) Equal to zero
Same Newton's-laws-of-motion concept applied to a car: at constant velocity the net force is zero (first law), whereas this NDA item computes the non-zero retarding force needed to decelerate the car (second law, F = ma).
- practice — not a real PYQ
A constant force of 20 N acts on a body of mass 4 kg. The acceleration produced is
- (a)80 m/s^2
- (b)5 m/s^2
- (c)24 m/s^2
- (d)0.2 m/s^2
Answer(b) 5 m/s^2 — a = F/m = 20/4.
- practice — not a real PYQ
A speed of 72 km/h is equal to
- (a)10 m/s
- (b)20 m/s
- (c)36 m/s
- (d)72 m/s
Answer(b) 20 m/s — divide km/h by 3.6.