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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Here's the solution to your physics problem: a) State Newton's second law of motion Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It is commonly expressed as F_net = ma, where F_net is the net force, m is the mass, and a is the acceleration. b) Two particles of mass 15 kg and 20 kg are connected by a string over a smooth light pulley. i) The common acceleration of the particles Step 1: Identify the forces acting on each mass. Let m_1 = 15 kg and m_2 = 20 kg. Let T be the tension in the string and g = 9.8 m/s^2 be the acceleration due to gravity. Since m_2 > m_1, m_2 will accelerate downwards and m_1 will accelerate upwards. For m_1 (15 kg, moving up): T - m_1 g = m_1 a (1) For m_2 (20 kg, moving down): m_2 g - T = m_2 a (2) Step 2: Add the two equations to eliminate T and solve for a. (T - m_1 g) + (m_2 g - T) = m_1 a + m_2 a m_2 g - m_1 g = (m_1 + m_2) a a = ((m_2 - m_1) g)/(m_1 + m_2) Step 3: Substitute the given values into the equation for a. a = (20 kg - 15 kg) × 9.8 m/s^215 kg + 20 kg a = (5 kg) × 9.8 m/s^235 kg a = 49 N35 kg a = 1.4 m/s^2 The common acceleration of the particles is 1.4 m/s^2. ii) The tension of the string Step 1: Substitute the calculated acceleration a into equation (1). T - m_1 g = m_1 a T = m_1 g + m_1 a T = m_1 (g + a) Step 2: Substitute the values for m_1, g, and a. T = 15 kg (9.8 m/s^2 + 1.4 m/s^2) T = 15 kg (11.2 m/s^2) T = 168 N The tension in the string is 168 N. c) A woman pulling her 20 kg suitcase at a constant speed. Given: Mass m = 20 kg, pulling force F_P = 35 N, frictional force F_f = 20 N. The suitcase moves at a constant speed, so the net acceleration is 0. i) What angle does the strap make with the horizontal? Step 1: Analyze the horizontal forces. Since the suitcase moves at a constant speed, the net horizontal force is zero. The horizontal component of the pulling force balances the frictional force. F_P - F_f = 0 F_P = F_f Step 2: Solve for . = (F_f)/(F_P) = 20 N35 N = (4)/(7) = ((4)/(7)) ≈ 55.15^ The angle the strap makes with the horizontal is 55.15^. ii) What normal force does the ground exert on the suitcase? Step 1: Analyze the vertical forces. Since there is no vertical acceleration, the net vertical force is zero. The normal force (F_N) and the vertical component of the pulling force (F_P ) balance the gravitational force (mg). F_N + F_P - mg = 0 F_N = mg - F_P Step 2: Substitute the values for m, g, F_P, and . F_N = (20 kg × 9.8 m/s^2) - (35 N × (55.15^)) F_N = 196 N - (35 N × 0.8206) F_N = 196 N - 28.72 N F_N = 167.28 N The normal force exerted by the ground is 167.28 N. iii) If the strap breaks, what would be the magnitude and direction of the acceleration of the suitcase? Step 1: Identify the forces acting on the suitcase after the strap breaks. When the strap breaks, the pulling force F_P becomes zero. The suitcase is still moving, so the frictional force F_f = 20 N will act opposite to the direction of motion. Step 2: Apply Newton's second law in the horizontal direction. The net force is the frictional force, acting opposite to the direction of motion. F_net = ma -F_f = ma a = -(F_f)/(m) Step 3: Calculate the acceleration. a = -20 N20 kg a = -1.0 m/s^2 The magnitude of the acceleration is 1.0 m/s^2. The negative sign indicates that the acceleration is in the direction opposite to the initial motion. The magnitude and direction of the acceleration are 1.0 m/s^2 opposite to the direction of motion. Send me the next one 📸