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Problems Involving Forces

Problems Involving Forces. Frictional Force. A smooth wooden block is placed on a smooth wooden tabletop. You find that you must exert a force of 14.0 N to keep the 40.0 N block moving at a constant velocity. a) What is the coefficient of sliding friction for the block and the table?

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Problems Involving Forces

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  1. Problems Involving Forces

  2. Frictional Force A smooth wooden block is placed on a smooth wooden tabletop. You find that you must exert a force of 14.0 N to keep the 40.0 N block moving at a constant velocity. a) What is the coefficient of sliding friction for the block and the table? b) If a 20.0 N brick is placed on the block, what force will be required to keep the block and brick moving at a constant velocity?

  3. Frictional Force FN REMEMBER: If the object is moving at a constant velocity, then the force applied (FA) will equal the Frictional force (Ff) FA Ff W Ff = µFN So µ = Ff/FN µ = 14.0 N/40.0 N µ = .350 Part a: Given: EQN: FA= Ff = 14.0 N Ff = µFN W = -40.0 N FN = 40.0 N µ = ?

  4. Frictional Force REMEMBER: If the object is moving at a constant velocity, then the force applied (FA) will equal the Frictional force (Ff) Part B: Ff = µFN so FA = µFN FA = (.350)(60.0 N) FA = 21.0 N New weight: W = -40.0 N + (- 20.0 N) = -60.0 N Therefore, the Normal Force: FN = 60.0 N Constant Velocity makes: FA = Ff

  5. Force and Acceleration A box weighs 75 N. a) What is the mass of the box? b) What is the acceleration of the box if an upward force of 90 N is applied?

  6. Force and Acceleration a) What is the mass of the box? Given: W = -75 N g = -9.8 m/s2 W = mg → m = W/g m = -75N/-9.8 m/s2 m = 7.7 kg

  7. Force and Acceleration b) What is the acceleration of the box if an upward force of 90 N is applied? Since the question is asking for acceleration, we need to use Newton’s 2nd Law, Fnet = ma Given: W = -75 N Upward force = 90 N m = 7.7 kg Fnet = ma → a = Fnet/m a = (-75 N + 90 N)/7.7 kg a = 15 N/7.7 kg a = 1.9 m/s2

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