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2.2 Friction questions
1. What is the maximum force of static friction between a block of ice that weighs 930 N and the ground if μs = 0.12? Ffr=μs FN FN = 930 N μs=0. Ffr=μs FN =( 930 ) ( 0.12)=111.6 N 2. What is the coefficient of static friction if it takes 34 N of force to start to move a box that has a mass of 7.0 kg? Ffr=μs FN Ffr= 34 N FN = 67 N μs=? 34 =μs ( 67 ) μs=0.5 1 3. A box takes 350 N to start moving when the coefficient of static friction is .35. What is the mass of the box? FN =? μs=0. Ffr= 34 N Ffr= 350 N Ffr=μk FN 350 =(0.35)F (^) N FN = 1000 N !!4. A car has a mass of 1020 kg and has a coefficient of static friction between the ground and its tires of 0.85. What force of friction can it exert on the ground? What is the maximum acceleration of this car? In what minimum distance could it stop from 27 m/s? F=ma=( 1020 kg ) ( 9.8)= 9996 N =FN Ffr=( 0.85 ) ( 9996 N ) =8496.6= 8500 N
F=ma 8496.6 N=( 1020 kg)( a) a=8.3 m/s x=? vi = 27 vf = 0 a=−8.33 m/s vf 2 =vi 2
- 2 ax 0 2 = 27 2
- 2 (−8.3 m/s ) s x= 44 m
5. A 0.170 kg hockey puck is initially moving at 21.2 m/s along the ice. The coefficient of kinetic friction for the puck and the ice is 0.005. (a) What is the speed of the puck after travelling 58.5 m? Fnet=Fk ma=μs FN
a=( 0.05) ( 0.17 kg ) (
9.8 m
s )
17 kg ¿ 0.049 m/s vf 2 =vi 2
- 2 ad ¿ ( 21.2) 2 −(0.049)(58.5) ¿ 21.13 m/s (b) After being played on for a while, the ice becomes rougher and the coefficient of kinetic friction increases to 0.047. How far will the puck travel if its initial and final speeds are the same as before? F=μk N F=0.047( 1.66) F=0.
7. A snowmobile is used to pull two sleds across the ice. The mass of the snowmobile and the rider is 320 kg. The mass of the first sled behind the snowmobile is 120 kg and the mass of the second sled is 140 kg. The ground exerts a force of 1500 N [forward] on the snowmobile. The coefficient of kinetic friction for the sleds on ice is 0.15. Assume that no other frictional forces act on the snowmobile. Calculate the acceleration of the snowmobile and sleds. Sled A: Normal Force f =ma f = 120 (9.8) f = 1176 N Friction Force F=μk N F=0.15( 1176 ) F=176.4 N Sled B Normal Force f =ma f = 140 (9.8) f = 1372 N Friction Force F=μk N F=0.15( 1372 )
F=208.5 N
Final Friction Fnet=208.5+176.4+ 1500 Fnet=1882. System Mass msyn= 320 + 120 + 140 msyn= 580 kg System Acceleration f =ma 188 2.2= 580 a a=3.2 m/s
8. A string is tied to a 3.2 kg object on a table and a 1.5 kg object hanging over a pulley. The coefficient of kinetic friction between the 3.2 kg object and the table is 0.30. (a) Calculate the acceleration of each object. (b) Determine the magnitude of the tension in the string. Mass A = 3.2kg Mass B = 1.5kg Coefficient Kinetic Friction between Mass A and table = 0. Total Mass: 3.2+1.5=4.7 kg Downward acceleration on Mass B: w=mg w=1.5(9.8) w=14.7 N Normal Force Mass A:
