University:
Concordia University Wisconsin
Course:
PHYS 1514 | General Physics I
Academic year:

2019
Views:

141

Pages:

2
Author:

gewoekerdodx

More Force Problems Name: (Key Date: 1. What force would be required to accelerate an 8.4 kg object at 2.7 m/s² with a force of friction of4.5N? (Answer: 27N) a 27 m 4.5N → ma-Vap-4.5N Fapp 8.4ty(27) +4.5N 27N 2. What is the mass of an object that accelerates from 15.0 m/s to 23.5 m/s over 12.0 seconds with an applied force of 32.0 N and a force of friction of 12.0 N? (Answer: 28.2kg) hushes V: 15 M 1125 a? a=Vr V 23.5-15=0.708 t NEI-NTE = YW M = 32-12 = 78.21g 3. A 20 kg box is sliding towards the right across the floor with an the blanks in the force diagram. acceleration of -1.0 m/s. Calculate and fill in all (Answer: Fx = 2.0 x 10%N: F-2.0x102N: F-200) J-98N FON F9 15 =-196N=-2.0X0N no applies furce problem a: -10m (slamice dayan foon friction) na Fr F= (2019-1.0m) Ave doction as ariel. (Answer: 61N) 4. What force would be required to accelerate a 5.0 kg mass from rest to a speed of 2.5 m/s, if the time of acceleration was only 0.30 s and a coefficient of friction of 0.40? Fout = Fapp-ff Fapp = Fru + = ma+uf =(518.3)+(0.4)(5X9.8) = 61.2N 6IN VLO €0.305 a = Ve-Vi t 0.3 8.3% Foff 5. A smooth wooden block is placed on a smooth wooden tabletop. You find that you must exert a force of 14.0N to keep the 40.0N block moving at a constant velocity-Fr a. What is the coefficient of sliding friction for the block and the table? (Answer: 0.350) 0 = 14N-M(40N) -14 με =0.350 b. If a 20.0N brick is placed on the block, what force will be required to keep the block and brick moving at a constant velocity? (Answer: 21.0N) app 0=Fopp-(0.350/60N) Key 6. A 2kg mass stretches a spring to 20cm. What mass will stretch the spring to 35cm? Answer: 3.5kg, F=KAX k = F = 2kgx9.8 N/ 20cm = 0.98N/cm F=KAX mg=10.98 (33cm) M = 0.9%-35 १.४ = 3.5kg 7. What is the force of friction acting on a 5.0 kg object that accelerates from an initial velocity of 2.0 m/s to a velocity of 8.45 m/s in 10.5 sec. The applied force acting on the object is 24.0 N. (Answer: -21N) a=? V=2% t=1955 a= V-V 815-2 = 0.614. Frut = Fapp-FF ma = Fapp-ff F = 24N-(5kg (0.6MB) = 20.9-21N ip direction to acceleration 8. Calculate the force of gravity between the Earth and a 70 kg satellite that is 4.56 x 10'm above the Earth's surface. Earth's radius = 6.38 x100m (Answer: 6.0 × 102N) GMm = (6.67x10 N² (5.98x15 (704) (6.38x10m+4.56×10) = 597.48N →6.0×10~ 9. What is the applied force acting on a 2.65 kg object that accelerates from rest to a velocity of 4.5 m/s in 2.5 seconds. The force of friction acting on the object is 4.35 Ν. (Answer: 9.1N) Fret = Fapp-Fr ma Fapp-4.35N (2.68kg) (18) +4.35N = Fapp Fap 9.1 N مله V=0 V=45M/S t=2.55 a=?= 4.5m/s. 2.5 1.8M/52 10. A 52N sled is pulled across a cement sidewalk at a constant speed. A horizontal force of 36N is exerted. a. What is the coefficient of sliding friction between the sidewalk and the metal runners of the sled? Cove spaa Fret = Fapp-F 36N-FF 3GN=MF 3GN=M(SAN) (Answer: 0.69) μ=0.69 b. Suppose the sled runs on packed snow. The coefficient of friction is now only 0.12. If a person weighing 650N sits on the sled, what force is needed to slide the sled across the snow at a constant speed? Fast=ma O app (Answer: 84N) A

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