Friction Problems and Analysis for Competitive Exams

Friction Problems and Analysis

Q. A 100 kg crate is resting on a horizontal surface. If the coefficient of static friction is 0.5, what is the maximum static friction force?

A. 490 N
B. 500 N
C. 510 N
D. 520 N

Solution

The maximum static friction force Ff = μs * N = 0.5 * (100 kg * 9.81 m/s²) = 490.5 N.

Correct Answer: B — 500 N

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Q. A 200 kg object is on a flat surface with a coefficient of kinetic friction of 0.3. What is the force of friction acting on the object?

A. 588 N
B. 600 N
C. 612 N
D. 620 N

Solution

The force of friction Ff = μk * N = 0.3 * (200 kg * 9.81 m/s²) = 588.6 N.

Correct Answer: A — 588 N

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Q. A 250 kg object is subjected to a horizontal force of 600 N. If the coefficient of friction is 0.5, will the object move?

A. Yes
B. No
C. Depends on the surface
D. Not enough information

Solution

The maximum static friction force Ff = μs * N = 0.5 * (250 kg * 9.81 m/s²) = 1226.25 N. Since the applied force (600 N) is less than the frictional force (1226.25 N), the object will not move.

Correct Answer: A — Yes

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Q. A block on an inclined plane has a mass of 10 kg and the angle of inclination is 30 degrees. What is the force of friction if the coefficient of kinetic friction is 0.2?

A. 5.88 N
B. 6.67 N
C. 7.84 N
D. 8.00 N

Solution

The normal force N = mg * cos(θ) = 10 kg * 9.81 m/s² * cos(30°) ≈ 84.87 N. The force of friction Ff = μk * N = 0.2 * 84.87 N ≈ 16.97 N.

Correct Answer: A — 5.88 N

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Q. A box weighing 300 N is pushed with a force of 150 N. If the coefficient of kinetic friction is 0.4, will the box move?

A. Yes
B. No
C. Depends on the surface
D. Not enough information

Solution

The force of kinetic friction Ff = μk * N = 0.4 * 300 N = 120 N. Since the applied force (150 N) is greater than the frictional force (120 N), the box will move.

Correct Answer: B — No

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Q. If a 150 kg object is on a slope of 15 degrees, what is the normal force acting on it?

A. 1430 N
B. 1500 N
C. 1600 N
D. 1700 N

Solution

The normal force N = mg * cos(θ) = 150 kg * 9.81 m/s² * cos(15°) ≈ 1430 N.

Correct Answer: A — 1430 N

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Q. If a car is moving at a speed of 20 m/s and the coefficient of friction between the tires and the road is 0.7, what is the maximum deceleration the car can achieve?

A. 14.7 m/s²
B. 19.6 m/s²
C. 9.81 m/s²
D. 7.0 m/s²

Solution

The maximum deceleration a car can achieve is given by a = μg, where g = 9.81 m/s². Thus, a = 0.7 * 9.81 m/s² ≈ 6.87 m/s².

Correct Answer: A — 14.7 m/s²

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Q. What is the angle of inclination at which the force of static friction equals the component of weight acting down the slope for a coefficient of static friction of 0.4?

A. 21.8 degrees
B. 22.5 degrees
C. 23.5 degrees
D. 24.0 degrees

Solution

The angle θ can be found using tan(θ) = μs. Thus, θ = arctan(0.4) ≈ 21.8 degrees.

Correct Answer: A — 21.8 degrees

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Q. What is the coefficient of static friction if a 50 kg box requires a force of 200 N to start moving?

A. 0.4
B. 0.5
C. 0.6
D. 0.7

Solution

The coefficient of static friction (μs) can be calculated using the formula F = μs * N, where N is the normal force. Here, N = mg = 50 kg * 9.81 m/s² = 490.5 N. Thus, μs = F/N = 200 N / 490.5 N ≈ 0.4.

Correct Answer: B — 0.5

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Friction Problems and Analysis

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