This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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You're on a roll — Here's the solution to question 16: 16. a) i) The coefficient of friction between the block and the turn table Step 1: Identify given values and convert units. Mass of block, m = 200 g = 0.2 kg Distance from center, r = 9 cm = 0.09 m Frictional force, F_f = 1.2 N Acceleration due to gravity, g = 9.8 m/s^2 Step 2: Calculate the normal force (N). The normal force on a horizontal surface is equal to the gravitational force. N = mg N = 0.2 kg × 9.8 m/s^2 N = 1.96 N Step 3: Calculate the coefficient of friction (_s). The maximum static frictional force is given by F_f = _s N. _s = (F_f)/(N) _s = 1.2 N1.96 N _s ≈ 0.6122 The coefficient of friction is 0.6122. 16. a) ii) The linear speed of the block Step 1: Relate frictional force to centripetal force. When the block begins to slide, the frictional force provides the centripetal force (F_c). F_c = F_f = 1.2 N Step 2: Use the formula for centripetal force in terms of linear speed. F_c = (mv^2)/(r) 1.2 N = 0.2 kg × v^20.09 m Step 3: Solve for the linear speed (v). v^2 = 1.2 N × 0.09 m0.2 kg v^2 = (0.108)/(0.2) v^2 = 0.54 m^2/s^2 v = sqrt(0.54 m)^2/s^2 v ≈ 0.7348 m/s The linear speed of the block is 0.7348 m/s. 16. a) iii) If the angular velocity is increased by 2 rad/s, what would be the force required to hold the block at the same place? Step 1: Calculate the initial angular velocity (_1) when the block begins to slide. The centripetal force is also given by F_c = mr^2. F_c = mr_1^2 1.2 N = 0.2 kg × 0.09 m × _1^2 1.2 = 0.018 _1^2 _1^2 = (1.2)/(0.018) _1^2 ≈ 66.6667 rad^2/s^2 _1 = sqrt(66.6667) ≈ 8.165 rad/s Step 2: Calculate the new angular velocity (_2). The angular velocity is increased by 2 rad/s. _2 = _1 + 2 _2 = 8.165 rad/s + 2 × 3.14159 rad/s _2 = 8.165 rad/s + 6.283 rad/s _2 ≈ 14.448 rad/s Step 3: Calculate the new force required (F_c2) to hold the block at the same place. F_c2 = mr_2^2 F_c2 = 0.2 kg × 0.09 m × (14.448 rad/s)^2 F_c2 = 0.018 × 208.7489 F_c2 ≈ 3.757 N The force required to hold the block at the same place is 3.757 N. 16. c) What is meant by the term "Banking" in roads? Banking in roads refers to the practice of raising the outer edge of a curved road or track relative to the inner edge. This tilt provides a component of the normal force that acts as a centripetal force, helping vehicles to negotiate turns safely at higher speeds without relying solely on friction. That's all for today. Send me the next one 📸