Dynamics - 3 - 2nd Law Against Gravity and Multi-Body

University:
High School
Course:
High School Physics
Academic year:

2024
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13
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 4.3 Working Against Gravity in nd 2 Law Recall: Frictional Forces: ⚫ ⚫ Act opposite to the direction of motion Types: ⚫ ⚫ ⚫ Static Friction Sliding Friction Rolling Friction ⚫ ⚫ Frictional Resistance ⚫ ⚫ Ex. Force between bike tire & road Ex. due to fluids/air/water Gravity is often an applied force, but can be a resistive force which must be overcome. Ex #1 When it is launched, a small rocket with a mass of 2.00 kg rises vertically with an acceleration of 12.0 m/s2. 2 12 m/s What force must the reactor exert to produce this acceleration? Fnet = Fapplied − Ffriction Fnet = ma Fnet = ( 2.00)(12.0) Fnet = 24.0 N FFric = ( 2.00)(9.80) FFric = 19.6 N 24.0 = Fapplied − 19.6 Fapplied = 43.6N Ex #2 An object of unknown mass is lifted with a vertical force of 18.0 N. Its acceleration as it rises is 2.00 m/s2. 18 N 2.0 m/s What is the object’s mass? 2 Fnet = Fapplied − Ffriction Fapplied = 18.0N Fnet = m(2.00) FFric = m(9.80) 2.00m = 18.0 − 9.80m 11.8m = 18.0 m = 1.53kg Ex #3 A 50.0 kg crate must be hoisted by rope onto an 8.00 m high platform. A person exerts 600.0N of force on a rope that goes through an overhead pulley. How long will it take to lift the crate from the ground to the platform? m = 50.0kg d = 8.00m Fapp = 600.0N v1 = 0m/s t=? FFric = (50.0)(9.80) FFric = 490N Fnet = Fapplied − Ffriction Fnet = 600 − 490 Fnet = 110N Fnet = ma 110 = (50.0)a a = 2.20m / s 2 1 d = v1t + at 2 2 1 8.00 = 0t + ( 2.20)t 2 2 t = 2.70s 4.4: Multi-Body Problems: ⚫ ⚫ To solve net force for a system of multiple connected objects, you can consider them as one single large object. All connected objects accelerate at the same rate! Ex. #4 A 10.0kg mass was suspended by a string & attached to a 5.00kg cart. The force of friction on the system was 40.0N. Mass 5kg net Friction = 40N applied 10kg What was the acceleration of the cart? Fnet = Fapplied − Ffriction Fapp = mg Fapp = (10.0)(9.80) Fapp = 98.0N Fnet = 98.0 − 40.0 Fnet = 58.0 N Fnet = ma 58.0 = (15.0)a a = 3.87m / s 2 Ex. #5 28.0 kg and 13.0 kg masses are attached to either end of a 25.0 kg cart. The force of friction between the table and cart is 75.0 N. Mass 25kg Friction = 75N 28kg 13kg m1 m2 What is the magnitude of the acceleration? Fnet = Fapplied − Ffriction Fapp = m1g Fnet = 274.4 − 202.4 Fapp = (28.0)(9.80) Fapp = 274.4 N Fnet = 72.0 N F fric = 75.0 + m2 g F fric = 75.0 + (13.0)(9.80) F fric = 202.4 N Fnet = ma (m total!) 72.0 = (66.0)a a = 1.09m / s 2