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Download Problems Involving Friction, Weight, and Elevators
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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? 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? 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 W Ff Ff = µFN Part a: Given: EQN: So µ = Ff/FN FA= Ff = 14.0 N Ff = µFN µ = 14.0 N/40.0 N W = -40.0 N FN = 40.0 N µ=? µ = .350 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: 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 Ff = µFN so FA = µFN FA = (.350)(60.0 N) FA = 21.0 N 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? 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 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
