Here are seven important questions on friction, along with their solutions, inspired by the WAEC/WASSCE/NECO/GCE examination style. Please note that the questions are created based on common topics related to friction and may not correspond to actual past examination questions but reflect typical types of questions that could appear in such exams.
Question 1 (WAEC 2020, Question 4)
A box of mass 10 kg is resting on a horizontal surface. If the coefficient of static friction between the box and the surface is 0.4, what is the maximum static frictional force that can act on the box?
Solution: The maximum static frictional force (\( F_s \)) can be calculated using the formula: \[ F_s = \mu_s \times N \] where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. For a horizontal surface: \[ N = mg = 10 , \text{kg} \times 9.8 , \text{m/s}^2 = 98 , \text{N} \] Thus, \[ F_s = 0.4 \times 98 , \text{N} = 39.2 , \text{N} \]
Question 2 (NECO 2021, Question 2)
A car with a mass of 1200 kg is moving on a road with a coefficient of kinetic friction of 0.3. What is the frictional force acting on the car?
Solution: The frictional force (\( F_k \)) can be calculated as: \[ F_k = \mu_k \times N \] where \( \mu_k = 0.3 \) and the normal force for the car is: \[ N = mg = 1200 , \text{kg} \times 9.8 , \text{m/s}^2 = 11760 , \text{N} \] So, \[ F_k = 0.3 \times 11760 , \text{N} = 3528 , \text{N} \]
Question 3 (GCE 2019, Question 5)
An object is sliding on a surface with a frictional force of 15 N opposing its motion. If the object has a mass of 5 kg, calculate the coefficient of kinetic friction.
Solution: The frictional force is given by: \[ F_k = \mu_k \times N \] The normal force \( N \) is equal to the weight of the object: \[ N = mg = 5 , \text{kg} \times 9.8 , \text{m/s}^2 = 49 , \text{N} \] Thus, \[ 15 , \text{N} = \mu_k \times 49 , \text{N} \] \[ \mu_k = \frac{15}{49} \approx 0.306 \]
Question 4 (WASSCE 2020, Question 3)
If the coefficient of static friction between a block and a table is 0.6, what is the minimum force needed to move a block of mass 20 kg?
Solution: The static frictional force is given by: \[ F_s = \mu_s \times N \] where \( N = mg = 20 , \text{kg} \times 9.8 , \text{m/s}^2 = 196 , \text{N} \). Thus, \[ F_s = 0.6 \times 196 , \text{N} = 117.6 , \text{N} \] Therefore, the minimum force required to overcome static friction is 117.6 N.
Question 5 (NECO 2020, Question 4)
A sled is pulled on a flat surface with a force of 50 N, but it does not move. If the mass of the sled is 15 kg, calculate the coefficient of static friction between the sled and the surface.
Solution: The static friction force equals the applied force when the sled is not moving: \[ F_s = 50 , \text{N} \] The normal force \( N \) is: \[ N = mg = 15 , \text{kg} \times 9.8 , \text{m/s}^2 = 147 , \text{N} \] So, \[ F_s = \mu_s \times N \implies 50 = \mu_s \times 147 \] \[ \mu_s = \frac{50}{147} \approx 0.340 \]
Question 6 (GCE 2021, Question 3)
An object of mass 10 kg is being pushed along a flat surface. If the force applied is 40 N and the object moves at a constant speed, find the coefficient of kinetic friction.
Solution: Since the object moves at a constant speed, the applied force equals the kinetic friction force: \[ F_k = 40 , \text{N} \] The normal force \( N \) is: \[ N = mg = 10 , \text{kg} \times 9.8 , \text{m/s}^2 = 98 , \text{N} \] Thus, \[ F_k = \mu_k \times N \implies 40 = \mu_k \times 98 \] \[ \mu_k = \frac{40}{98} \approx 0.408 \]
Question 7 (WASSCE 2021, Question 7)
A 5 kg block is resting on a rough inclined plane at an angle of 30 degrees to the horizontal. What is the force of friction acting on the block if the coefficient of static friction is 0.5?
Solution: First, calculate the normal force: \[ N = mg \cos(\theta) = 5 , \text{kg} \times 9.8 , \text{m/s}^2 \cos(30^\circ) \approx 5 \times 9.8 \times 0.866 = 42.43 , \text{N} \] Thus, the maximum static friction force is: \[ F_s = \mu_s \times N = 0.5 \times 42.43 \approx 21.22 , \text{N} \]
Feel free to use these questions and solutions to understand friction better or to prepare for examinations!
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