Laws of Motion Study Guide for Competitive Exams

Laws of Motion

Circular Motion Friction Newtons Laws

Q. A force of 15 N is applied to a 3 kg object. What is the resulting acceleration?

A. 3 m/s²
B. 5 m/s²
C. 7 m/s²
D. 10 m/s²

Solution

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

Correct Answer: B — 5 m/s²

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Q. A force of 15 N is applied to a 5 kg object. What is the object's acceleration?

A. 3 m/s²
B. 2 m/s²
C. 1 m/s²
D. 4 m/s²

Solution

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

Correct Answer: A — 3 m/s²

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Q. A force of 30 N is applied to a 5 kg object. What is the object's acceleration?

A. 3 m/s²
B. 6 m/s²
C. 9 m/s²
D. 12 m/s²

Solution

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

Correct Answer: B — 6 m/s²

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Q. A mass m is attached to a string and is whirled in a horizontal circle. If the radius of the circle is halved, what happens to the tension in the string if the speed remains constant?

A. It doubles
B. It remains the same
C. It halves
D. It quadruples

Solution

Tension T = mv²/r. If r is halved, T doubles for constant speed.

Correct Answer: A — It doubles

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Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, what is the minimum speed required to keep the mass in circular motion?

A. √(g*r)
B. g*r
C. 2g*r
D. g/2

Solution

At the highest point, the centripetal force is provided by the weight. Minimum speed = √(g*r).

Correct Answer: A — √(g*r)

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Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, the tension in the string is T. What is the expression for T?

A. T = mg
B. T = mg - mv²/r
C. T = mg + mv²/r
D. T = mv²/r

Solution

At the highest point, T + mg = mv²/r, thus T = mg - mv²/r.

Correct Answer: B — T = mg - mv²/r

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Q. A mass m is attached to a string and is whirled in a vertical circle. At the highest point of the circle, what is the condition for the mass to just complete the circular motion?

A. Tension = 0
B. Tension = mg
C. Tension = 2mg
D. Tension = mg/2

Solution

At the highest point, the centripetal force is provided by the weight of the mass, so T + mg = mv²/r. For T = 0, mg = mv²/r.

Correct Answer: A — Tension = 0

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Q. A mass m is attached to a string and is whirled in a vertical circle. At the top of the circle, the tension in the string is T. What is the expression for the tension at the bottom of the circle?

A. T + mg
B. T - mg
C. T
D. T + 2mg

Solution

At the bottom, T + mg = mv²/r; T = mv²/r - mg.

Correct Answer: A — T + mg

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Q. A mass m is attached to a string of length L and is swung in a vertical circle. At the highest point of the circle, what is the minimum speed required to keep the mass in circular motion?

A. √(gL)
B. √(2gL)
C. gL
D. 2gL

Solution

At the highest point, the centripetal force must equal the weight: mv²/L = mg, thus v = √(gL).

Correct Answer: A — √(gL)

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Q. A particle moves in a circle of radius 3 m with a speed of 6 m/s. What is the angular velocity of the particle?

A. 1 rad/s
B. 2 rad/s
C. 3 rad/s
D. 4 rad/s

Solution

Angular velocity ω = v/r = 6 m/s / 3 m = 2 rad/s.

Correct Answer: B — 2 rad/s

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Q. A particle moves in a circular path of radius 10 m with a constant speed of 5 m/s. What is the period of the motion?

A. 2π s
B. 4π s
C. 10 s
D. 20 s

Solution

Circumference = 2πr = 20π m. Period T = distance/speed = 20π m / 5 m/s = 4π s.

Correct Answer: A — 2π s

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Q. A particle moves in a circular path of radius 10 m with a speed of 5 m/s. What is the time period of the motion?

A. 2π s
B. 4π s
C. 10 s
D. 20 s

Solution

Time period T = 2πr/v = 2π(10 m) / (5 m/s) = 4π s.

Correct Answer: A — 2π s

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Q. A particle moves in a circular path of radius 5 m with a constant speed of 10 m/s. What is the centripetal acceleration of the particle?

A. 2 m/s²
B. 5 m/s²
C. 10 m/s²
D. 20 m/s²

Solution

Centripetal acceleration (a_c) = v²/r = (10 m/s)² / (5 m) = 20 m/s².

Correct Answer: C — 10 m/s²

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Q. A particle moves in a circular path of radius r with a constant angular acceleration α. What is the expression for the angular displacement θ after time t?

A. θ = αt²
B. θ = 0.5αt²
C. θ = αt
D. θ = 0.5αt

Solution

Angular displacement θ = 0.5αt² for constant angular acceleration.

Correct Answer: B — θ = 0.5αt²

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Q. A particle moves in a circular path of radius r with a constant angular speed ω. What is the time period of the motion?

A. T = 2πr/ω
B. T = ω/2πr
C. T = 2π/ω
D. T = r/ω

Solution

The time period T = 2π/ω for uniform circular motion.

Correct Answer: C — T = 2π/ω

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Q. A particle moves in a circular path with a constant speed. Which of the following statements is true?

A. The particle has zero acceleration
B. The particle has constant acceleration
C. The particle has centripetal acceleration
D. The particle's velocity is constant

Solution

The particle has centripetal acceleration directed towards the center of the circle.

Correct Answer: C — The particle has centripetal acceleration

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Q. A particle moves in a circular path with a radius of 10 m and completes one revolution in 5 seconds. What is the linear speed of the particle?

A. 2π m/s
B. 4π m/s
C. 10 m/s
D. 20 m/s

Solution

Linear speed (v) = Circumference / Time = (2π * 10 m) / 5 s = 4π m/s.

Correct Answer: A — 2π m/s

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Q. A particle moves in a circular path with a radius of 10 m at a constant speed of 5 m/s. What is the period of the motion?

A. 2π s
B. 4π s
C. 10 s
D. 20 s

Solution

Circumference = 2πr = 20π m. Time = distance/speed = 20π m / 5 m/s = 4π s.

Correct Answer: B — 4π s

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Q. A particle moves in a circular path with a radius of 15 m and completes one revolution in 10 seconds. What is the linear speed of the particle?

A. 1.5 m/s
B. 3 m/s
C. 6 m/s
D. 9 m/s

Solution

Linear speed (v) = 2πr/T = 2π(15 m)/10 s = 3 m/s.

Correct Answer: B — 3 m/s

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Q. A particle moves in a circular path with a radius of 2 m and completes one revolution in 4 seconds. What is the linear speed of the particle?

A. 1.57 m/s
B. 3.14 m/s
C. 6.28 m/s
D. 12.56 m/s

Solution

Circumference = 2πr = 2π(2) = 4π m. Speed = Distance/Time = 4π/4 = π m/s ≈ 3.14 m/s.

Correct Answer: C — 6.28 m/s

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Q. A particle moves in a circular path with a radius of 4 m and completes one revolution in 8 seconds. What is the centripetal acceleration?

A. 0.5 m/s²
B. 1 m/s²
C. 2 m/s²
D. 4 m/s²

Solution

Centripetal acceleration (a_c) = v²/r. First find v = 2πr/T = 2π(4)/8 = π m/s. Then a_c = (π)²/4 = 2.5 m/s².

Correct Answer: B — 1 m/s²

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Q. A person is pushing a heavy box across a floor. If the box does not move, what can be said about the forces acting on it?

A. Net force is zero
B. Net force is positive
C. Net force is negative
D. Only friction is acting

Solution

If the box does not move, the net force acting on it is zero, meaning the applied force equals the frictional force.

Correct Answer: A — Net force is zero

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Q. A person pushes a box with a force of 30 N, but the box does not move. If the coefficient of static friction is 0.6, what is the maximum static friction force?

A. 18 N
B. 30 N
C. 36 N
D. 60 N

Solution

The maximum static friction force is equal to the applied force when the box does not move, which is 30 N.

Correct Answer: C — 36 N

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Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is doubled, what happens to the gravitational force acting on the satellite?

A. It doubles
B. It halves
C. It becomes four times
D. It becomes one-fourth

Solution

Gravitational force ∝ 1/r². If radius is doubled, force becomes 1/(2²) = 1/4.

Correct Answer: D — It becomes one-fourth

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Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is increased, what happens to the orbital speed of the satellite?

A. Increases
B. Decreases
C. Remains the same
D. Depends on the mass of the satellite

Solution

Orbital speed (v) = √(GM/r). As r increases, v decreases.

Correct Answer: B — Decreases

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Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is 7000 km and the gravitational acceleration is 9.8 m/s², what is the speed of the satellite?

A. 5.5 km/s
B. 7.9 km/s
C. 9.8 km/s
D. 11.2 km/s

Solution

Speed (v) = √(g*r) = √(9.8 m/s² * 7000 * 1000 m) = 7.9 km/s.

Correct Answer: B — 7.9 km/s

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Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is 7000 km and the speed of the satellite is 7.9 km/s, what is the centripetal acceleration?

A. 7.9 m/s²
B. 9.8 m/s²
C. 11.2 m/s²
D. 14.0 m/s²

Solution

Centripetal acceleration a_c = v²/r = (7.9 km/s)² / (7000 km) = 0.00063 km/s² = 6.3 m/s².

Correct Answer: A — 7.9 m/s²

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Q. A satellite is in a circular orbit around the Earth. If the radius of the orbit is tripled, how does the orbital speed change?

A. It triples
B. It doubles
C. It remains the same
D. It decreases by a factor of √3

Solution

Orbital speed v = √(GM/r). If r is tripled, v decreases by a factor of √3.

Correct Answer: D — It decreases by a factor of √3

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Q. A stone is thrown in a circular path with a radius of 3 m. If the stone completes 4 revolutions in 8 seconds, what is the angular speed?

A. π/2 rad/s
B. π rad/s
C. 2π rad/s
D. 4π rad/s

Solution

Angular speed (ω) = Total angle / Time = (4 * 2π) / 8 = π rad/s.

Correct Answer: B — π rad/s

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Q. A stone is tied to a string and swung in a vertical circle. At the highest point, the tension in the string is 5 N and the weight of the stone is 10 N. What is the speed of the stone at the highest point if the radius of the circle is 2 m?

A. 2 m/s
B. 3 m/s
C. 4 m/s
D. 5 m/s

Solution

At the highest point, T + mg = mv²/r. 5 + 10 = (m*v²)/2. Solving gives v = 4 m/s.

Correct Answer: C — 4 m/s

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Laws of Motion

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