Laws of Motion

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Q1. An object moves with a constant velocity of 10 m/s. What is the net force acting on it? (2023)

Solution:

If the object moves with constant velocity, the net force acting on it is 0 N.

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Q2. If a 10 kg object is dropped from a height of 20 m, what is the velocity just before it hits the ground? (Assume g = 10 m/s²) (2019)

Solution:

Using v² = u² + 2gh, v = √(0 + 2 × 10 m/s² × 20 m) = √400 = 20 m/s.

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Q3. A force of 30 N is applied to a 6 kg object. What is the object's acceleration? (2023)

Solution:

Using F = ma, we find a = F/m = 30 N / 6 kg = 5 m/s².

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Q4. A 50 kg object is moving with a velocity of 5 m/s. What is its momentum? (2023)

Solution:

Momentum p = mv = 50 kg * 5 m/s = 250 kg·m/s.

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Q5. If a force of 15 N is applied to a mass of 3 kg, what is the acceleration of the mass? (2023)

Solution:

Using F = ma, a = F/m = 15 N / 3 kg = 5 m/s².

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Q6. A ball is thrown vertically upwards with a speed of 20 m/s. How high will it go? (g = 10 m/s²) (2023)

Solution:

Using the formula h = v²/(2g), we have h = (20 m/s)² / (2 * 10 m/s²) = 20 m.

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Q7. A 2 kg object is subjected to a net force of 8 N. What is its acceleration? (2020)

Solution:

Using F = ma, a = F/m = 8 N / 2 kg = 4 m/s².

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Q8. A cyclist accelerates at 3 m/s² for 4 seconds. What is the final velocity if the initial velocity is 2 m/s? (2023)

Solution:

Using v = u + at, we have v = 2 m/s + (3 m/s² * 4 s) = 2 + 12 = 14 m/s.

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Q9. A 2 kg ball is thrown vertically upwards with a velocity of 20 m/s. What is the maximum height reached? (2020)

Solution:

Using v² = u² + 2as, 0 = (20 m/s)² + 2(-9.8 m/s²)(h). Solving gives h = 20.4 m.

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Q10. A 50 kg object is pushed with a force of 200 N. What is the frictional force if the object moves with constant velocity? (2023)

Solution:

If the object moves with constant velocity, the frictional force equals the applied force, which is 200 N.