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Transcript
Physics Practice Exam Solutions February 14th 1. [A] Magnitude of A- B = √[(Ax-Bx)² + (Ay-By)²]=√[(-2.9-3.4)² + (7.0-(-4.2))²]= 12.85 2. [B] Magnitude of vector = √[(-4)²+(5)²+(10)²] = √141 Now find x, y, and z components in ratios to the magnitude, which gives you the unit vector, so i= -4/√141=-0.33, j=5/√141=0.42, k=10/√141=0.84 3. [B] If you draw a free-body diagram, you have: sum of F=ma, if she weighs 1000N, then her mass is 1000N/(9.8m/s²) =102.04 kg, and her acceleration is 4.9 m/s², so F=ma=(102.04 kg) • (4.9 m/s²) = 500 N 4. [D] Using conservation of energy, we come up with the equation: 0.5kx²=mgh. Solving for k, we get k=2mgh/x²=2(.180 kg)(9.8 m/s²)(1 m) / (0.20)² = 88.2 N/m 5. [D] In drawing a diagram of where John walks, you will find he is 1 km east from where he started, so this is his total displacement. The total time is (3 km)/(2.5 km/hr) + (2 km)/(15.0 km/hr) + (20 min) • (1 hr) / (60 min) + (30 min) • (1 hr) /(60 min) = 2.16667 hr. So the average velocity is (1 km)/( 2.16667 hr) = 0.46 km/hr 6. [E] If we have v(t), we can find a(t) just by taking a derivative, so a(t)= 3at² + 4at³ = 3t² + 8t³, so a(3)= 3(3)² + 8(3)³ = 243 7. [D] If you draw a straight line from City A to City B, this will be the sum of your two vectors, the plane and the wind. If you put the plane and wind tip to tail with the sum being the line between the cities, then you have a right triangle and you solve for the cities vector: √(210²-150²)= 146.97 ( km/hr). Since the cities are 600 km apart, then 600 km / (146.97 km/hr) = 4.08 hr 8. [E] Studying the forces when the block stops moving on the incline, the maximum static friction force is fs=μsN=(0.6)(1)(9.8)cos(30)=5.097 N, and the gravity pulling the block down the ramp is mgsinθ=(1)(9.8)sin(30)= 4.905 N. The static friction force is stronger than the gravity pulling the block down the ramo, so the block will never return to the bottom of the ramp. 9. [B] Sum of F= msysasys=15N, solving for asys: asys=15N/(50 kg) = 0.3 m/s² 10. [C] We can find the F on each block, FA= (30 kg)(0.3 m/s²)= 9 N, and FB= (20 kg)(0.3 m/s²)= 6 N. FA/ FB=9 N / 6 N = 1.5 11. [D] Taking along the string as a positive axis, the sum of F= msysasys= 3mg. Now we can solve for asys: asys=(3mg)/(6m) = g/2 12. [C] You can draw a free body diagram at the top of the hill, and there is a gravity and then the circular force as the car is moving in circular motion, in a cross-section sense. Sum of F=0=mv²/r - mg, solving for v, we get v=√(gr)=√[(9.8 m/s²)(1024 m)]= 100.18 m/s 13. [C] Just as in Paula’s Lecture 10, you need to find the net force of the system: 60 N – 40 N =20 N. Now find the acceleration of the system: asys=Fnet/msys=(20 N)/(10 kg) = 2 m/s². The sum of the forces on the string is: ma=FR-T, where m is the mass of the blocks pulling on the string in the direction of motion, so T=FR-ma=60-4(2)=44N 14. [B] We need to find when static friction force equals the magnitude of the force keeping the ball in circular motion, so when ma= μsmg, so when a= μsg, then μs=a/g. There are two components of a, a radial and a tangential. So arad=rω² and mrω² 3.04 m/s², so μs=3.04/9.8=0.31. 15. [B] We can find the displacement, θ, in radians, by the equation ωf²-ω0²=2αθ, which is a simple conversion from the 1-dimensional equation to circular motion. Solving for θ=(4.00)²/(2 • 0.02)= 40 rad. Now, we just make a conversion from rad to revolutions: 40 rad • [(1 rev)/(2π rad)] = 6.4 rev 16. [A] This is a conservation of energy problem, setting it up, we get: mgh+0.5mv²=0.5kx², because there is both potential and kinetic energy initially, then it is all elastic potential at the end. Solving for x, we get: x=√[(mgh+0.5mv²)/(0.5k)]=√[((5)(9.8)(2)+(0.5)(5)(10)²)/((0.5)(1000))]=0.834 m Or, by using work-kinetic energy, the same numbers are also found. 17. [D] Using conservation of energy, in order to avoid some error to rounding, we can use the total energy of the initial system: mgh2=mgh1+0.5mv², solving for h2, we get h2=[(gh1+0.5v²)/g]=[(9.8•2 + 0.5•10²)/9.8]=7.10 m 18. [C] There are a few ways to do this problem. The way I would do it is using conservation of energy: mgh + 0.5mv0²=0.5mvf², solving for h: h=[(0.5vf²0.5v0²)/(g)]=[(0.5(92)²-0.5(45)²)/9.8]=330 m You can use projectile motion equations, also. We can find the y component if velocity and solve for the height. Since vx=45cos30=38.97, and is constant for a projectile, the final magnitude of velocity is 92, so 922=vx2+vy2, vy=√(922-38.972) = 83.34. Using vf2=v02+2ad, where d is h, then h=(83.342-(45sin30)2)/(2*9.8)=328m. 19. [D] Using W=F•d=Fdcosθ, we have W=250•4•cos36=809 N. 20. [E] First find the sum of the two forces, which make a right angle: Sum of F = √(FJ²+FS²)=√(100²+144²)= 175 N. The block is 1 kg, and a=F/m=175/10= 1.75 m/s²