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Transcript
Physics Review By: Kellie Krueger Velocity, Acceleration, and Displacement Velocity V l i – Rate R off change h off position. ii Change in position / change in time ∆x/∆t vf² = vi² + 2a∆x vf = at + vi Acceleration – increase or decrease in velocity velocity. Change in velocity / change in time ∆v/∆t Displacement – the vector quantity that defines the distance and direction between 2 positions. ∆x = ½ at2 + vit Example Problem A bike travels at a constant speed of 4.0 m/s for 5 s. How far does it go? 1. v = ∆x/∆t 2. 4 0m/s = ∆x / 5s 4.0m/s 3. 4.0(5s) 4. ∆x = 20m Example Problem 2 A car can accelerate from rest to speed of 28m/s in 20s. a) what is the average acceleration of the car? a = ∆v/∆t a = 28m/s / / 20s a = 1.4m/s / 2 Example Problem 3 A 65 kg person dives into the water from a 10m platform. What is her speed as she enters the water? Given: Mass = 65 kg therefore Fg = -650N ∆x = 10m Trying to find: Speed Force Diagram Normal Force 65 kg Force of gravity = -650N Solving the Problem vf² = vi² + 2a∆x vf² = (0m/s)² + 2(9.8m/s²)(10m) vf = 14m/s Force, Mass, and Acceleration The sum of the forces is equal to mass multiplied by acceleration. F = ma Example Problem An elevator is moving up at a constant velocity of 2.5m/s. The mass of the man = 85 kg. What force does the floor exert on the man? Given: G e Because of constant velocity (2.5m/s), a = 0m/s² Mass M = 85kg, 85k therefore, th f Fg = -850N 850N Trying to find: Normal Force Force Diagram Normal Force Gravity Force = -850N Solving the Problem F = ma -850N = (85kg)(0m/s²) Normall Force = 850N Newton’s 2nd Law and Friction Coefficient of friction: µ FF= µ ٭FN Example Problem A sled weighing 300N is moved at a constant speed over a horizontal floor by a force of 50N applied parallel to the floor. What is the coefficient of friction between the sled and the floor? Force Diagram Normal Force: 300N Friction Force: ? Applied Force: 50N Gravity Force: -300N Givens and Unknowns Given: Since there is a constant speed, p ,a=0 Applied force = 50N Gravity force = -300N 300N Normal force = 300N Trying to find: Coefficient of friction Solving the Problem µ = FF/FN µ = 50N/300N µ = .167 Problem 2 What would be the acceleration of the sled if µ = 0? F = ma 50N = 30kg(a) a = 1.67 m/s²