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Chapter 4.3-4.4 Review

Chapter 4.3-4.4 Review. 1. If an 8 kg object is accelerating at a rate of 20 m/s 2 , what is the net force on the object?. F = ma F = 8 x 20 F = 160 N. 2. If you applied the same force as the answer to question 1 to a 4 kg object, what acceleration would result?.

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Chapter 4.3-4.4 Review

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  1. Chapter 4.3-4.4 Review

  2. 1. If an 8 kg object is accelerating at a rate of 20 m/s2, what is the net force on the object?

  3. F = maF = 8 x 20F = 160 N

  4. 2. If you applied the same force as the answer to question 1 to a 4 kg object, what acceleration would result?

  5. F = ma160 = 4 x aa = 40 m/s2

  6. 3. What are the action reaction pairs in these examples? A) a bat hits a baseball B) wind causes a kite to fly C) the moon orbits the earth D) you turn a steering wheel in your car E) you use a remote to change channels on your TV

  7. A) a bat hits a baseballThe bat pushes on the ball, the ball pushes on the bat.

  8. B) wind causes a kite to fly The air pushes on the kite, the kite pushes on the air.

  9. C) the moon orbits the earth The earth pulls on the moon, the moon pulls on the earth.

  10. D) you turn a steering wheel in your car Your hand pushes on the wheel, the wheel pushes on your hand.

  11. E) you use a remote to change channels on your TVYour finger pushes on the button, the button pushes on your finger.

  12. 4. Newton’s third law says that for every force applied, an equal and opposite force results. If that is true, why don’t the forces cancel each other out? How can any force produce acceleration?

  13. The forces are not on the same object, they are on two different objects. Each of these forces COULD be unbalanced and COULD cause acceleration.

  14. 5. The forces on a motorcycle are 470 N north and 250 N west. If the motorcycle and rider have a combined mass of 210 kg, what is the magnitude and direction of the acceleration?

  15. 4702 + 2502 = c2c = 532 Ntanq = 470/25062°north of west

  16. F = ma532 = 210 x aa= 2.5 m/s262°north of west

  17. 6. A bug flying east at 5 m/s hits the windshield of a car moving west at 20 m/s. (He won’t have the guts to do that again!) Which experiences the greater impact force, the bug or the windshield? Which experiences the greater acceleration? Why?

  18. The impact forces are equal (Newton’s Third Law). The bug has a greater acceleration because it has a smaller mass (F = ma).

  19. 7. Draw a free body diagram of each of these cases: A) a thrown baseball (without air resistance) B) a thrown baseball (with air resistance) C) a grocery cart being pushed D) a bottle rocket E) a skydiver falling at terminal velocity

  20. A. Only weight. This is an example of projectile motion.

  21. B. Weight and air resistance, only.

  22. Assuming the cart is not accelerating, it must be in equilibrium. FN = mg and Ff = FP.

  23. I assume the rocket is still firing and accelerating. Thrust is greater than air resistance; the only other force is gravity.

  24. Since the skydiver has reached terminal velocity, the air resistance is equal to the gravity. The skydiver is in equilibrium and is not accelerating.

  25. 8. A box has a mass of 4.7 kg. What is its weight on earth? What is its weight on the moon, where g is 1/6 that on earth?

  26. Wt = mgWt = 4.7 x 10Wt = 47 NWt = mgWt = 4.7 x 10 x 1/6Wt = 7.8 N

  27. 9. A wooden block is on an incline of 33º. Friction is holding it in place. What are the magnitudes of the frictional force and the normal force?

  28. The components of the weight are the force parallel to the plane and the force perpendicular to the plane.sin33° = Fparallel /mg Fparallel = 0.545mg Nparallelcos33° = Fperpendicular /mg Fperpendicular = 0.839mg Nperpendicular

  29. Since the object is in equilibrium, the frictional force is equal to the parallel force.Fparallel = Ffrictional = 0.545mg NSince the object is in equilibrium, the frictional force is equal to the parallel force.Fperpendicular = Fnormal = 0.839mg N

  30. 10. A hockey puck on the ice is at rest. It has a mass of 1.5 kg. A force of 11 N is required to start the puck moving. After that a force of 9 N is required to keep the puck moving at a constant velocity. What are the coefficients of static and kinetic friction?

  31. Since the ice is flat, the normal force is equal to the weight, mg.m = Ffriction/Fnormalm = 11 N/(1.5 x 10) Nm = 0.733

  32. Since the ice is flat, the normal force is equal to the weight, mg.m = Ffriction/Fnormalm = 9 N/(1.5 x 10) Nm = 0.600

  33. 11. Can the force of air resistance on a falling body exceed the force of gravity?

  34. No, the object would accelerate until air resistance was equal to weight. At that point the object would be in equilibrium, and it would no longer accelerate. (terminal velocity)

  35. The next slide is extra! Don’t read it unless you understand the previous answer!

  36. (Couldn’t stand it could you?)Actually, since the density of air increases as you approach the surface of the earth, the air resistance WOULD increase above the pull of gravity and slow the speed of terminal velocity as the object approaches the earth’s surface. But, that would only be appreciable if the object fell for a LONG distance.

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