Table 1: List of Equations for Law of Acceleration Second Law of Motion Fema If acceleration is missing but force and mass is given: If mass is missing but force and acceleration is given: m = = If two objects are presented: Fi Acceleration: a = Av At - = " or a = = VI Problem Solving Sample Problem 1. How much force is needed to accelerate a 2,500kg cargo truck at a rate of 3m/s?? 2. What is the mass of a box that requires a force of 40 N to accelerate at a rate of 5m/s?? 1. Solution: 2. Solution Given: Give: Mass (m) = 2,500kg Force (F) = 40 N Acceleration (a) = 3m/s Acceleration (a) = 5m/s? Formula: F=ma Formula m= - m = 40 N F = (2,500kg)( 3m/s?) 5 m/s? F = 7,50ON m = 8kg Practice Problem 1. What is the acceleration of an object if it ha as mass of 540kg and a force of 8N is acted on it? 2. An object accelerates at 5.8m/s with a force of 82 N. An object with similar mass accelerates at 8.5 m/s'. How much force was exerted on the second object? 3. Find the force given to an object that travels for 1 minute with an initial velocity of 12 m/s and a final velocity of 18.2m/s. The mass of the object is 3,829 kg.Problem Solving Sample Problem 1. A . 5.5 kg tennis ball moved at 7m/s. What is its kinetic energy? 2. Compare which among the 2 balls have a higher kinetic energy. A red ball with 5,000g that travels at 7m/s and a blue ball that has a mass of 7kg that travels at 6 m/s 1. Solution 2. Solution Given Given m = 5.5kg v = 7m/s mi = 5,000g = 5kg v1 = 7m/s m2 = 7kg Formula V2 = 6m/s Formula myz KE = 2 KE = (5.5kg) (7m/s)2 Red ball KE = 2 KE = (5kg)(7m/s)2 2 KE = 134.75 ] KE = 122.5] Blue ball (7kg)(6m/s)? KE = 2 KE = 126 J The blue ball has the higher KE Practice Problem 1. Find the KE of a 5,745g object that has a velocity of 17m/s. 2. Who has the greatest KE among the 3 animals. Their average velocities are as follows; a lion that travels 22.35m/s, a tiger with 17.88m/s and a human with 13.21m/s. Their average mass is 190kg. 170kg, and 62kg respectively.Problem Solving Sample Problem The pitcher threw the ball horizontally from 5.5m with an initial speed of 25 m/s. a) How long will it take the ball to reach the ground? b) What is the range of the c) What is the magnitude of its velocity when it strikes the ground? c) What angle will it strike the ground? Given: h = 5.5m (-5.5m) g = 9.8m/s2 v = 25m/s Solution: Get the x and y component: Vix = Vi . COSO 25m 25m Vix = -. COSO = S Vly = v . sine 25m Viy . sin0 = 0m/s S a) Find the time. Use the y component value. dy = vnit+-gt2 1 -5.5 m = 0+ = (-9.8m/s2) t2 anspose 0 + 1/2 (9.8m/s?) to the left to derive the value to 12 then get the square root. (2) (-5.5 m) 9.8 m/s2 V12 = V1. 12 52 1 = 1.06 s b) Range or horizontal component displacement. R = vil R = (25 %) (1. 065) R = 26. 5m c) use the Pythagorean Theorem to get the magnitude v = v, 2 + vy Get the vy component to get the magnitude using this formula. Vy = Viy + gt Vy = 0+ ( - - 9.8my $2 (1.065) = -10.39m/s with all values complete substitute the values to solve for v v = V (25 m/s2)2 + (-10.39 m/s2)2 v = V625m/s2 + (-107.95 m/s2) v = V517.05) v = 22. 74m/s d) Angle 0 = tan -1 = tan -1 25m/ s 1-10.39m/S = tan '|-0.421 = 22.78 Practice Problem The pitcher threw the ball horizontally from 6.88m with an initial speed of 35 m/s. a) How long will it take the ball to reach the ground? b) What is the range of the c) What is the magnitude of its velocity when it strikes the ground? c) What angle will it strike the ground