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Unformatted text preview: Practice Exam 2 1. The mass of the Moon is about 1/80th of the mass of Earth. The force exerted by Earth on the Moon is about 80 times that exerted by the Moon on Earth. a. True b False Solution: Since Newton’s 3rd Law states the “every action has an equal and opposite reaction”, the forces experienced by the Earth and Moon will be equal and opposite. (b. False) 2. The force of kinetic friction between two surfaces is dependent on the relative speed of the two surfaces. a. True b. False Solution: Kinetic friction is calculated by the following: ඃ" = ඃ" ඃ& None of the components of the formula are dependent on speed. False 4. A force of 120N is applied to an object whose mass is 30 kg. The object’s acceleration is a. 4.0 ඃ.  ඃ b. 3600 ඃ.  ඃ c. 0.25 ඃ.  ඃ d. 2.0 ඃ.  ඃ e. 150 ඃ.  ඃ Solution: 4 ඃ5 = ඃ = ඃඃ 120ඃ = (30ඃඃ)ඃ ඃ. ඃ ඃ. ඃ = ඃ ඃ 3. You ride on an elevator that is moving downward with constant speed while standing on a bathroom scale. The reading on the scale is a. Equal to your true weight b. Less than your true weight c. More than your true weight d. Could be more or less than your true weight, ඃඃ, depending on the value of the speed. Solution: Constant speed ⟹ no acceleration ⟹ ∑ ඃ = 0 4 ඃD = ඃ& − ඃඃ = 0 ඃ& = ඃඃ The reading on the scale is ඃ& which equals mg or your true weight. 5. A catcher stops a ball traveling at 40 m/s in a distance of 20cm and feels a force of 600N against his glove. What is the mass of the ball? a. 0.15 kg b. 0.20 kg c. 0.25 kg d. 0.10 kg e. 0.30 kg Solution: ඃ = ∆ඃ −ඃඃ = 1 1 ඃඃK  − ඃඃL  2 2 1 −ඃඃ = − ඃඃL  2 2ඃඃ 2(600ඃ)(0.20ඃ) = ඃ = = ඃ. ඃඃඃඃ (40 ඃ⁄ඃ)ඃL 6. Three boxes rest side by side on a smooth horizontal floor. Their masses are 5.0kg, 3.0kg, and 2.0kg, with the 3.0kg mass in the center. A force of 50N pushes on the 5.0kg mass, which pushes against the other two masses. What is the contact force between the 5.0kg mass and the 3.0kg mass? a. 50N b. 25N c. 10N d. 40N e. 0N Solution: First look at the system as a whole. 4 ඃ5 = ඃ + (ඃST − ඃTS ) + (ඃT − ඃT ) = (ඃS + ඃT + ඃ )ඃ ඃ = (ඃU + ඃ + ඃT )ඃ ඃ 50ඃ = ඃ = = 5 ඃ.  ඃ (ඃS + ඃT + ඃ ) 10ඃඃ Now look just at block 5. 4 ඃ5 = ඃ − ඃTS = ඃS ඃ ඃ − ඃS ඃ = ඃTS 50ඃ − (5ඃඃ) V5 ඃ.  W = ඃඃඃ ඃ 5 5 3 2 7. The two forces indicated in Figure 1 act on a 3.00kg object. What is the acceleration of the object? a. V5.00 ඃ.  W ඃ + V1.00 ඃ.  W ඃ Z Z ඃ ඃ b. V1.67 ඃ.  W ඃ − V0.333 ඃ.  W ඃ Z Z ඃ ඃ c. V1.67 ඃ.  W ඃ + V2.333 ඃ.  W ඃ Z Z ඃ ඃ d. V1.67 ඃ.  W ඃ + V0.333 ඃ.  W ඃ Z Z ඃ ඃ e. V15.0 ඃ.  W ඃ + V3.00 ඃ.  W ඃ Z Z ඃ ඃ Solution: First look at the system as a whole. 5 3 2 4 ඃ5 = 4ඃ + 1ඃ = 5ඃ = (3.0ඃඃ)ඃ5 5ඃ = ඃ. ඃඃ ඃ. ඃ ඃ 3.0ඃඃ 4 ඃ_ = 4ඃ − 3ඃ = 1ඃ = (3.0ඃඃ)ඃD 1ඃ = ඃ. ඃඃඃ ඃ. ඃ ඃ 3.0ඃඃ 8. A 1000‐kg barge is being towed by means of two horizontal cables. One cable is pulling with a force of 80.0 N in a direction 30.0° west of north. The second cable pulls in a direction 20.0° east of north. What should the magnitude of its pulling force be so that the barge will accelerate northward? a. 117N b. 58.5N c. 120N d. 127N e. 73.7N Solution: 4 ඃ5 = ඃ cos 70° − 80ඃ cos 60° = 0 ඃ cos 70° = 80ඃ cos 60° ඃ = 80ඃ cos 60° = 117ඃ cos 70° 9. A policeman investigating an accident measures the skid marks left by a car. He determines that the distance between the point that the driver slammed on the brakes and the point where the car came to a stop was 28.0 m. From a reference manual he determines that the coefficient of kinetic friction between the tires and the road under the prevailing conditions was 0.300. How fast was the car going when the driver applied the brakes? (This car was not equipped with anti‐lock brakes.) a. 10.7m/s b. 12.8m/s c. 21.4m/s d. 45.7m/s e. 32.9m/s Solution: Use the conservation of energy equation: ∆ඃ + ∆ඃ = −ඃ" ඃ 1 1 g ඃඃK  − ඃඃL  h + (0 − 0) = −ඃ" ඃ& ඃ 2 2 1 − ඃඃL  = −ඃ" ඃ& ඃ 2 1 ඃඃL  = ඃ" ඃඃඃ 2 ඃL  = 2ඃ" ඃඃ ඃL = i2ඃ" ඃඃ = ඃඃ. ඃ ඃ⁄ඃ 10. A mass of 40.0 grams is attached to a vertical spring with a spring constant k = 20.0 N / m and lowered slowly until the spring stops stretching. How much does the spring stretch? a. 0.0816m b. 0.200m c. 0.00200m d. 0.800m e. 0.0196m Solution: 4 ඃD = ඃk − ඃඃ = 0 ඃk = ඃඃ ඃඃ = ඃඃ ඃඃ = ඃ. ඃඃඃඃඃ ඃ 11. A plane has an airspeed of 142 m/ s. A 16.0 m/ s wind is blowing southward at the same time as the plane is flying. If the velocity of the plane relative to Earth is due east, what is the magnitude of that velocity? a. 141m/s b. 16.2m/s c. 158m/s d. 16.0m/s e. 48m/s ඃ = Solution: Using tail to tip method ඃmnop + ඃqrsot = ඃ (ඃmnop ) + ඃ  = vඃqrsot w ඃ = xvඃqrsot w − (ඃmnop ) = ඃඃඃ ඃ⁄ඃ  12. A 600‐kg car is going around a curve with a radius of 120m that is banked at an angle of 20° with a speed of 24.5 m/ s. What is the minimum coefficient of static friction required for the car not to skid? a. 0.36 b. 0.60 c. 0.24 d. 0.12 e. 0.48 Solution: NOTE: Dr. Stancil’s solution is INCORRECT! 4 ඃ5 = ඃ& cos ඃ − ඃඃ − ඃk sin ඃ = 0 ඃ& cos ඃ − ඃk ඃ& sin ඃ = ඃඃ ඃ& (cos ඃ − ඃk sin ඃ ) = ඃඃ ඃ& = 4 ඃD = ඃ& sin ඃ + ඃk cos ඃ = ඃඃ ඃ& sin ඃ + ඃk ඃ& cos ඃ = ඃඃ ඃ& (sin ඃ + ඃk cos ඃ ) = ඃඃ ඃ& = ඃඃ sin ඃ + ඃk cos ඃ ඃඃ cos ඃ − ඃk sin ඃ Now set the two equations equal to each other: ඃඃ ඃඃ = cos ඃ − ඃk sin ඃ sin ඃ + ඃk cos ඃ ඃඃ(sin ඃ + ඃk cos ඃ ) = ඃඃ (cos ඃ − ඃk sin ඃ ) ඃ sin ඃ + ඃඃk cos ඃ = ඃ cos ඃ − ඃ ඃk sin ඃ ඃඃk cos ඃ + ඃ ඃk sin ඃ = ඃ cos ඃ − ඃ sin ඃ ඃk (ඃ cos ඃ + ඃ sin ඃ ) = ඃ cos ඃ − ඃ sin ඃ ඃ cos ඃ − ඃ sin ඃ 1.35 ඃ cos ඃ − ඃ sin ඃ ඃ ඃk = = = = ඃ. ඃඃඃ ඃ ඃ cos ඃ + ඃ sin ඃ 10.9 ඃ cos ඃ + sin ඃ ඃ 13. Two masses are connected by a string which goes over an ideal pulley as shown in Figure 2. Block A has a mass of 3.0 kg and can slide along a smooth plane inclined 30° to the horizontal. What is the mass of block B if the system is in equilibrium? a. 1.5kg b. 6.0kg c. 2.6kg d. 3.0kg e. 3.0kg Solution: 4 ඃ5 = ඃ − ඃ€ ඃ sin ඃ = 0 ඃ = ඃ€ ඃ sin ඃ 4 ඃD = ඃ − ඃ• ඃ = 0 ඃ = ඃ• ඃ Since the pulley is ideal, the tension, ඃ is same on either side of the pulley. ඃ = ඃ€ ඃ sin ඃ = ඃ• ඃ ඃ€ sin ඃ = ඃ• = ඃ. ඃඃඃ 14. What would be the weight of a 59.1‐kg astronaut on a planet with the same density as Earth and having twice Earth's radius? a. 290N b. 1160N c. 580N d. 1200N e. 2320N Solution: ඃඃඃඃඃඃඃ†s‡ˆ‰ = ඃqrsotˆ ඃඃඃඃ ඃ†s‡ˆ‰ = = 4 ඃඃඃඃඃඃ 4 ඃඃ T T ඃඃ 3 †s‡ˆ‰ 3 qrsotˆ = ඃqrsotˆ 4 ඃ(2ඃ†s‡ˆ‰ )T 3 ඃ†s‡ˆ‰ 4 T ඃඃ 3 †s‡ˆ‰ ඃqrsotˆ 8 8ඃ†s‡ˆ‰ = ඃqrsotˆ Weight on Earth is ඃඃ. Weight on planet: ඃඃඃqrsotˆ ඃඃ(8ඃ†s‡ˆ‰ ) ඃඃqrsotˆ = = (2ඃ†s‡ˆ‰ )ඃqrsotˆ 8 ඃ (ඃ†s‡ˆ‰ ) ඃ (ඃ†s‡ˆ‰ ) ඃqrsotˆ = =2 = 2ඃ (ඃ†s‡ˆ‰ )4 (ඃ†s‡ˆ‰ ) ඃඃqrsotˆ = 2ඃඃ = ඃඃඃඃඃ ඃ†s‡ˆ‰ = 15. How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/ s to 40.0 m/ s? a. 20.0kJ b. 1.10MJ c. 0.960MJ d. 40.0kJ e. 10.0kJ Solution: 1 1 ∆ඃ = g ඃඃK  − ඃඃL  h = 1.10 × 10_ ඃ = ඃ. ඃඃඃඃ 2 2 ඃK = 40 ඃ⁄ඃ , ඃL = 15 ඃ⁄ඃ 16. You are trying to cross a river that flows due south with a strong current. You start out in your motorboat on the east bank desiring to reach the west bank directly west from your starting point. You should head your motorboat? a. In a northwesterly direction b. Due west c. Due south d. Due north e. In a southwesterly direction Solution: In a northwesterly direction 17. An object of 1.0 kg mass is pulled up an inclined plane by a constant force of 10 N that causes a displacement of 0.50 m. The angle of inclination with the horizontal is 30°. Neglect friction and use g = 10 m/s2. What is the work done by the 10 N force on the object along the inclined plane? a. 4.3J b. 4.3kgඃ.  c. 2.5J d. 5.0J ඃ Solution: ඃ = ඃ ∆ඃ = 10ඃ(0.50ඃ) = ඃ. ඃඃ 18. The hydrogen atom consists of a proton of mass 1.67 × 10”• ඃඃ and an orbiting electron of mass 9.11 × 10”TU ඃඃ. In one of its orbits, the electron is 5.3 × 10”UU ඃ from the proton. What is the mutual attractive gravitational force between the electron and proton? a. 9.3 × 10”–• ඃ b. 3.6 × 10”–• ඃ c. 5.4 × 10”–• ඃ d. 1.8 × 10”–• ඃ e. 7.0 × 10”–• ඃ Solution: ඃ ඃq ඃt ඃ = = ඃ. ඃ × ඃඃ”ඃඃ ඃ ඃ 19. A 5.00‐kg box slides 4.00 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/ s? a. 0.587 b. 0.229 c. 0.115 d. 1.13 e. 0.267 Solution: Use the conservation of energy equation: ∆ඃ + ∆ඃ = −ඃ" ඃ 1 1 g ඃඃK  − ඃඃL  h + (0 − 0) = −ඃ" ඃ& ඃ 2 2 1 − ඃඃL  = −ඃ" ඃ& ඃ 2 1 ඃඃL  = ඃ" ඃඃඃ 2 1 ඃඃL 2 = ඃ. ඃඃඃ ඃඃඃ 20. The ratio of kinetic energy of object A to the kinetic energy of object B is 2:1. Object A is moving with a speed of 6.0 m/ s and object B is moving with a speed of 2.0 m/ s. If the mass of object A is 4.0 kg then what is the mass of object B? a. 2.0kg b. 16kg c. 8.0kg d. 18kg e. 4.0kg Solution: 1 (4ඃඃ)(6ඃ/ඃ)ඃ€ 2 2 ඃ€ ඃ€ == = ඃ• 1 1 ඃ ඃ ඃ• (2ඃ/ඃ)2 •• 144 2= 4ඃ• 8ඃ• = 144 ඃ• = ඃඃඃඃ ...
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This note was uploaded on 04/16/2011 for the course PHYS 1111 taught by Professor Plascak during the Spring '08 term at University of Georgia Athens.
 Spring '08
 plascak
 Force, Mass

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