1. Practice with Kinetic Friction Renate Fiora

2. Let’s try solving a problem involving kinetic friction. Remember the equation for the force of friction: F f =  F N Don’t forget to include a drawing and free- body diagram.

3. Bonnie and Clyde are sliding a 300. kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 385 N of force while Bonnie pulls forward on a rope with 350 N of force. What is the safe’s coefficient of kinetic friction on the bank floor? Try it on your own, then advance to the next slide to see the solution.

4. Bonnie and Clyde are sliding a 300. kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 385 N of force while Bonnie pulls forward on a rope with 350 N of force. What is the safe’s coefficient of kinetic friction on the bank floor? Because there’s no net force, we can say:  F y = F N – F W = 0  F N = F W = mg  F x = F B + F C – F f = 0  F B + F C = F f So,  = F f /F N = (F B + F C )/mg = (350N + 385N)/(300kg)(9.8 m/s 2 )  =  0.25 a = 0 FWFW FfFf FBFB FNFN FCFC +x +y FWFW FNFN F net = 0 FBFB FfFf FCFC

5. A man slides a 45 g crate at a constant velocity across a horizontal floor by pulling on a rope attached to the crate. The angle  between the rope and horizontal is 33  and the coefficient of kinetic friction between crate and floor is 0.63. Determine the tension in the rope. Try it on your own, then advance to the next slide to see the solution.

6. A man slides a 45 kg crate at a constant velocity across a horizontal floor by pulling on a rope attached to the crate. The angle  between the rope and horizontal is 33  and the coefficient of kinetic friction between crate and floor is 0.63. Determine the tension in the rope. There’s no net force, so we can say:  F y = F N + F Ty – F W = 0 F N + F Ty = F W = mg F N = mg - F T sin   F x = F Tx – F f = 0 F T cos  = F f So, F f =  F N F T cos  =  mg -  F T sin  F T (cos  +  sin  =  mg F T =  mg/(cos  +  sin  a = 0 FWFW FfFf FTFT FNFN +x +y FWFW FNFN F net = 0 FTFT FfFf   Now, plug in the values: F T =  /(cos  +  sin   F T = 