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Practice Finland Ylioppilastutkinto Physics with 100 original study MCQs covering core lukio knowledge, application, and exam strategy. Official test is a 6-hour digital paper (up to 7/11 tasks, max 120 points).

Sample Physics Practice Questions

Try these sample questions to test your Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A car travels at constant velocity 20 m/s for 15 s. How far does it travel?
A.300 m
B.35 m
C.150 m
D.400 m
Explanation: Distance = speed × time for constant velocity: s = 20 m/s × 15 s = 300 m.
2An object starts from rest and accelerates uniformly at 3.0 m/s² for 4.0 s. What is its final speed?
A.7.0 m/s
B.12 m/s
C.1.5 m/s
D.48 m/s
Explanation: v = v₀ + at = 0 + (3.0 m/s²)(4.0 s) = 12 m/s.
3A ball is dropped from rest. After 2.0 s of free fall near Earth (g = 9.8 m/s², air resistance neglected), its speed is closest to:
A.9.8 m/s
B.40 m/s
C.20 m/s
D.4.9 m/s
Explanation: v = gt = 9.8 × 2.0 = 19.6 m/s ≈ 20 m/s.
4What is the SI unit of force?
A.watt (W)
B.pascal (Pa)
C.newton (N)
D.joule (J)
Explanation: Force has SI unit newton: 1 N = 1 kg·m/s² from F = ma.
5A 2.0 kg object is acted on by a net force of 10 N. What is its acceleration?
A.5.0 m/s²
B.8.0 m/s²
C.0.20 m/s²
D.20 m/s²
Explanation: Newton's second law: a = F/m = 10 N / 2.0 kg = 5.0 m/s².
6Work done by a constant force of 50 N moving an object 3.0 m in the direction of the force is:
A.150 J
B.50 J
C.300 J
D.17 J
Explanation: W = F s cos0° = 50 N × 3.0 m = 150 J.
7Kinetic energy of a 4.0 kg mass moving at 3.0 m/s is:
A.18 J
B.6.0 J
C.36 J
D.12 J
Explanation: KE = ½mv² = 0.5 × 4.0 × 9.0 = 18 J.
8Gravitational potential energy of a 5.0 kg mass raised 2.0 m (g = 9.8 m/s²) is closest to:
A.10 J
B.196 J
C.49 J
D.98 J
Explanation: PE = mgh = 5.0 × 9.8 × 2.0 = 98 J.
9Momentum of a 0.50 kg ball moving at 8.0 m/s is:
A.16 kg·m/s
B.4.0 kg·m/s
C.8.5 kg·m/s
D.0.50 kg·m/s
Explanation: p = mv = 0.50 × 8.0 = 4.0 kg·m/s.
10A box is pulled at constant speed on a rough floor. The net force on the box is:
A.zero
B.equal to the friction force only
C.equal to its weight
D.greater than the applied force
Explanation: Constant velocity means zero acceleration, so by Newton's first/second law the net force is zero: applied force balances friction.

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