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 are the solutions to the questions. 4.1.1 State Newton's Second Law of motion in words. Newton's Second Law of motion states that the net force acting on an object is directly proportional to the object's mass and its acceleration, and acts in the same direction as the acceleration. 4.1.2 Draw a labelled free body diagram showing all the forces acting on the car. A free body diagram for the car moving up an inclined surface at an angle would show the following forces originating from the center of the car: • Weight (F_g or W): Acts vertically downwards. • Normal force (F_N or N): Acts perpendicular to the inclined surface, pointing upwards. • Tension (T): Acts parallel to the inclined surface, pointing upwards along the cable. • Kinetic frictional force (F_f or f_k): Acts parallel to the inclined surface, pointing downwards (opposite to the direction of motion). 4.1.3 Calculate the kinetic frictional force that the car experiences as it moves up the surface. Given: Mass of car (m_car) = 800 kg Coefficient of kinetic friction (_k) = 0.25 Angle of incline () = 30^ Acceleration due to gravity (g) = 9.8 m/s^2 Step 1: Calculate the normal force (F_N). The normal force on an inclined plane is the component of the gravitational force perpendicular to the surface. F_N = m_car g F_N = (800 kg)(9.8 m/s^2) (30^) F_N = 7840 N × 0.8660 F_N = 6790.99 N Step 2: Calculate the kinetic frictional force (F_f). The kinetic frictional force is given by the product of the coefficient of kinetic friction and the normal force. F_f = _k F_N F_f = (0.25)(6790.99 N) F_f = 1697.75 N The kinetic frictional force is 1698 N. 4.1.4 Calculate the tension T in the cable. Given: Mass of car (m_car) = 800 kg Mass of truck (m_truck) = 2500 kg Net force on truck (F_net, truck) = 1520 N Angle of incline () = 30^ Acceleration due to gravity (g) = 9.8 m/s^2 Kinetic frictional force (F_f) = 1697.75 N (from 4.1.3) Step 1: Calculate the acceleration of the system. Since the truck is pulling the car with a cable, they both experience the same acceleration. a = F_net, truckm_truck a = 1520 N2500 kg a = 0.608 m/s^2 Step 2: Apply Newton's Second Law to the car along the incline. The net force on the car along the incline is the sum of the forces acting parallel to the incline. F_net, car = T - m_car g - F_f According to Newton's Second Law, F_net, car = m_car a. m_car a = T - m_car g - F_f Step 3: Solve for tension (T). T = m_car a + m_car g + F_f T = (800 kg)(0.608 m/s^2) + (800 kg)(9.8 m/s^2) (30^) + 1697.75 N T = 486.4 N + (7840 N)(0.5) + 1697.75 N T = 486.4 N + 3920 N + 1697.75 N T = 6104.15 N The tension in the cable is 6104 N. 4.2.1 State Newton's Law of universal gravitation in words. Newton's Law of Universal Gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. 4.2.2 Calculate the distance between the spaceship and the centre of Earth when Earth exerts a force of 2458 N on the spaceship. Given: Force (F) = 2458 N Mass of spaceship (m_spaceship) = 1250 kg Gravitational constant (G) = 6.67 × 10^-11 N·m^2/kg^2 Mass of Earth (M_E) = 5.972 × 10^24 kg Step 1: Write down the formula for universal gravitation. F = G m_spaceship M_Er^2 Step 2: Rearrange the formula to solve for the distance (r). r^2 = G m_spaceship M_EF r = sqrt(G m_spaceship) M_EF Step 3: Substitute the given values and calculate r. r = sqrt((6.67 × 10^-11) N·m^2/kg^2)(1250 kg)(5.972 × 10^24 kg)2458 N r = sqrt(4.978355 × 10^17) N·m^22458 N r = sqrt(2.02536 × 10^14) m^2 r = 1.42315 × 10^7 m The distance between the spaceship and the center of Earth is 1.42 × 10^7 m. Send me the next one 📸