1 / 17

Q05. Using Newtons Laws

Q05. Using Newtons Laws. A 13- N weight and a 12- N weight are connected by a massless string over a massless, frictionless pulley. The 13- N weight has a downward acceleration with magnitude equal to that of a freely falling body times: 1 1/12 1/13 1/25 13/25. T. .

lmacias
Télécharger la présentation

Q05. Using Newtons Laws

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Q05. Using Newtons Laws

  2. A 13-N weight and a 12-N weight are connected by a massless string over a massless, frictionless pulley. The 13-N weight has a downward acceleration with magnitude equal to that of a freely falling body times: • 1 • 1/12 • 1/13 • 1/25 • 13/25

  3. T  T a 12 N a 13 N

  4. A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If the coefficients of friction are s = 0.5 and k = 0.4, the magnitude of the frictional force on the crate is: • 8 N • 12 N • 16 N • 20 N • 40 N

  5. 12 N f s = 0.5 k = 0.4 40 N • crate is stationary •  f= 12 N.

  6. A bureau rests on a rough horizontal surface (s = 0.50, k = 0.40). A constant horizontal force, just sufficient to start the bureau in motion, is then applied. The acceleration of the bureau is: • 0 • 0.98 m/s2 • 3.9 m/s2 • 4.9 m/s2 • 8.9 m/s2

  7. F m g fk s = 0.50 k = 0.40

  8. A car is traveling at 15 m/s on a horizontal road. The brakes are applied and the car skids to a stop in 4.0 s. The coefficient of kinetic friction between the tires and road is: 0.38 0.69 0.76 0.92 1.11

  9. A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static friction is s. To start the crate moving with the weakest possible applied force, in what direction should the force be applied? Horizontal sin1s below the horizontal tan1s above the horizontal sin1s above the horizontal cot1s above the horizontal

  10. F  m g fs 

  11. A professor holds an eraser against a vertical chalkboard by pushing horizontally on it. He pushes with a force that is much greater than is required to hold the eraser. The force of friction exerted by the board on the eraser increases if he: • pushes with slightly greater force • pushes with slightly less force • stops pushing • pushes so his force is slightly downward but has the same magnitude • pushes so his force is slightly upward but has the same magnitude

  12. F m g fs • pushes with slightly greater force fs unchanged • pushes with slightly less force fs unchanged • stops pushing fs = 0 • pushes so his force is slightly downward but has the same magnitude • fs = m g + Fy • pushes so his force is slightly upward but has the same magnitude • fs = m g  Fy

  13. A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.70 and the coefficient of kinetic friction is 0.50. After one end of the plank is raised so the plank makes an angle of 30° with the horizontal, the force of friction is: • 13 N • 18 N • 22 N • 25 N • 30 N

  14. s = 0.70 k = 0.50 5.0 kg   = 30 

  15. A block is suspended by a rope from the ceiling of a car. When the car rounds a 40-m radius horizontal curve at 10 m/s, what angle does the rope make with the vertical? • 0° • 14 °  • 45 °  • 76 °  • 90 ° 

  16. T m v2 / R m g  0° 14 ° 45 ° 76 ° 90 °

More Related