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Practice Problems (p.21-27)

Practice Problems (p.21-27). To go from this…………………………...to this. . Impulse. An object experiences a force of 19.97 N for a time period of 4.58 s. What is the impulse of the object?. Solution. J=F( t). J= 19.97 (4.58) . Try another one! .

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Practice Problems (p.21-27)

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  1. Practice Problems (p.21-27) To go from this…………………………...to this.

  2. Impulse An object experiences a force of 19.97 N for a time period of 4.58 s. What is the impulse of the object?

  3. Solution J=F( t) J= 19.97 (4.58)

  4. Try another one! For what time must you exert a force of 45 N to get an impulse of 16 Ns?

  5. Solution 16=45(x) X= .36

  6. Momentum A 142.5 kilogram motorcycle is moving at a speed of 67.5 m/s. What is the momentum of the cycle?

  7. Solution P=mv P=142.5(67.5)

  8. Try another one! What is the momentum of a 23 Kg cannon shell going 530 m/s?

  9. Solution P=mv P=23(530) P= 12,190

  10. Conservation of Momentum Judy (mass=40.0 kg) is standing on slippery ice and catches her leaping dog, Atti (mass=15 kg), moving horizontally at 3.0 m/s. What is the speed of Judy and her dog after the catch?

  11. Solution (40)(0) + (15)(3)= mJudyi(vJudyi)+ mAtti(vAtti)= mJudyf(vJudyf)+ mAttif(vAttif) 40 kg 15 kg = 55 kg V= 3m/s P=45 (15x3) V= 45=55(v) .82 m/s P=45 V= 0 m/s P= 0 P=0+45 P=45

  12. Try another one! A 120 kg lineman moving west at 2 m/s tackles an 80 kg football fullback moving east at 8 m/s. After the collision, both players move east at 2 m/s. What is their final momentum?

  13. Solution 80 kg 120 kg 200 kg V= 8 m/s P= 640 V= -2 m/s P=-240 V=2 m/s P= 400 P= 640+-240 =400

  14. Amplitude/ Period Amplitude Period

  15. Hooke’s Law 1. A force of 600 Newtons will compress a spring 0.5 meters. What is the spring constant of the spring?2. A spring has spring constant 0.1 m/Newton. What force is necessary to stretch the spring by 2 meters?

  16. Solution 600= -.5 (x) =-1200 F= -.1(2) =-0.2 FHooke’s=-k( x)

  17. Try another one! What is the force required to stretch a spring whose constant value is 100 N/m by an amount of 0.50 m?

  18. Solution FHooke’s=-k( x) F= 100(.50) F=50 N

  19. Torque In a hurry to catch a cab, you rush through a frictionless swinging door and onto the sidewalk. The force you exerted on the door was 50N, applied perpendicular to the plane of the door. The door is 1.0m wide. Assuming that you pushed the door at its edge, what was the torque on the swinging door (taking the hinge as the pivot point)?

  20. Solution T=(1.0m) (50N) T= 50Nm

  21. Try another one! A force of 20 N is applied perpendicular to the end of a bar of length 0.5 m. Calculate the torque produced by the force.

  22. Solution T= (0.5)(20N) T=10Nm

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