- Basic Math and Science Newton's Laws of Motion Questions and Answers Flashcards

A 5 kg block is pushed along a frictionless horizontal surface by a 20 N force at an angle of 30° below the horizontal. What is the acceleration of the block?

Answer: 3.46 m/s²

Only the horizontal component of the applied force causes horizontal acceleration. The horizontal component is F·cos(30°) = 20 × 0.866 = 17.32 N. Using Newton's Second Law: a = F_x / m = 17.32 / 5 = 3.46 m/s². The vertical component (F·sin(30°) = 10 N downward) increases the normal force but does not contribute to horizontal acceleration on a frictionless surface.

Two blocks, A (3 kg) and B (5 kg), are connected by a massless string and pulled across a frictionless surface by a 24 N force applied to block B. What is the tension in the string connecting A and B?

Answer: 9 N

The total mass of the system is 3 + 5 = 8 kg. The acceleration is a = F / m_total = 24 / 8 = 3 m/s². The tension in the string only needs to accelerate block A: T = m_A × a = 3 × 3 = 9 N. This illustrates that the string transmits only the force needed to pull the trailing block, not the entire applied force.

A person stands on a scale inside an elevator. The scale reads 20% less than the person's true weight. Which of the following best describes the elevator's motion?

Answer: Accelerating downward at 0.2g

The apparent weight equals m(g − a) when the elevator accelerates downward. If the scale reads 80% of true weight: m(g − a) = 0.8mg, which gives g − a = 0.8g, so a = 0.2g downward. Note that decelerating upward and accelerating downward produce identical scale readings because both result in a net downward acceleration, but 'accelerating downward at 0.2g' is the more precise description of the net effect.

A 10 kg object is in equilibrium on a frictionless inclined plane tilted at 30°. A horizontal force F is applied to keep it stationary. What is the magnitude of F?

Answer: 56.6 N

For equilibrium on a frictionless incline with a horizontal force, the normal force N acts perpendicular to the surface and must balance both the weight component perpendicular to the plane and the horizontal force component. Summing forces: vertically, N·cos30° = mg → N = mg/cos30°. Horizontally, F = N·sin30° = (mg/cos30°)·sin30° = mg·tan30° = 10 × 9.8 × 0.577 ≈ 56.6 N. This is a non-trivial equilibrium problem because the normal force direction changes the calculation.

An astronaut has a mass of 70 kg. On a planet where the gravitational acceleration is 4 m/s², the astronaut pushes off a 200 kg spacecraft with a force of 140 N for 2 seconds. What is the speed of the spacecraft after the push, assuming it starts from rest?

Answer: 1.4 m/s

By Newton's Third Law, the spacecraft experiences an equal and opposite force of 140 N. Using Newton's Second Law for the spacecraft: a = F/m = 140/200 = 0.7 m/s². After 2 seconds: v = a × t = 0.7 × 2 = 1.4 m/s. The planet's gravitational acceleration and the astronaut's mass are distractors — what matters is the force on the spacecraft and its mass. In space with no friction, gravity direction doesn't affect the horizontal push.

A 2 kg block rests on a 5 kg block, which sits on a frictionless table. The coefficient of static friction between the two blocks is 0.4. What is the maximum horizontal force that can be applied to the lower block so that both blocks accelerate together without slipping?

Answer: 27.44 N

The only horizontal force on the upper block (2 kg) is static friction from the lower block. The maximum static friction on the upper block is f_s = μ_s × m_upper × g = 0.4 × 2 × 9.8 = 7.84 N. This gives the maximum acceleration: a_max = f_s / m_upper = 7.84 / 2 = 3.92 m/s². Applying this to the entire system: F_max = (m_upper + m_lower) × a_max = (2 + 5) × 3.92 = 27.44 N. A common error is applying the force limit only to one block rather than the combined system.