KINS 382 LAB 14

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KINS 382 - LAB 14

• Work and Power

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Question 1

• Under what circumstances will the work done by a
  force (F) be
• a. positive?
• b. negative?

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Answer 1

• a. Positive work is done when the direction of
  the displacement is the same as the direction of
  the force.
• b. Negative work occurs when the displacement is
  in the opposite direction of the applied force.

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Question 2

• Chris Bitcon sleeps through class without moving.
  Is the work done on him by the force of gravity
  positive, negative or zero?
• Explain!

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Answer 2

• Zero There is no work being done by gravity
  because there is no displacement of the object
  that the force is being applied to.

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Question 3

• An 80kg pole vaulter falls 5m from the crossbar
  into the landing pit.
• a. How much work is done on the athlete by
  gravity?
• Explain why the answer is positive.

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Answer 3

• a. W F x d or W (m)(a) x d
• a. W 80kg(-9.8m/s²) x -5m
• a. W 784 (kg)m/s² or N x -5m
• a. W 3920 Nm or Joules J
• b. Work is because the displacement is in the
  same direction of the force.

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Question 4

• Jeremy (mass 70kg) jumps into a pool from a 10m
  diving platform. Calculate the work done on him
  by gravity during the fall.

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Answer 4

• W F x d
• W (m)(a) x d
• W 70kg (-9.8m/s²) x -10m
• W 686 (kg) m/s² or N x -10m
• W 6860 Nm or Joules J

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Question 5

• A rope climber (mass 80kg) climbs 10 meters in
  5 seconds. In the process what power does he
  develop? (Velocity is constant).

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Answer 5

• P F x V
• P W/t
• P F x d/t
• Find V, V 10m / 5s 2m/s
• Find F, F 80kg X 9.8N/1kg 784 N
• P F x V
• P 784N (2m/s)
• P 1568 Nm/s or J/s 1568 Watts

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Question 6

• An athlete weighing 900N climbs 12 meters at a
  speed of 1.2m/s. During the climb the athlete
  generates how much power?

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Answer 6

• F 900N, d 12m, V 1.2 m/s
• P F x V
• P 900N (1.2m/s)
• P 1080 Nm/s
• P 1080 Watts

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Question 7

• A swimmer moves through the water at a constant
  speed of 0.2m/s. The drag force opposing her
  motion is 100 N.
• a. What is the power developed by the drag force?
• b. Explain why it is negative?
• c. At 0.4m/s the drag force is 400N. What power
  is developed at this speed?

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Answer 7

• V 0.2m/s, F -100N, P ?
• a. P F x V
• a. P -100N X 0.2m/s
• a. P -20 Nm/s or -20 Watts

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Answer 7 (Cont)

• b. Drag force is negative because it opposes the
  direction of the displacement.

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Answer 7 (Cont)

• c. V 0.4m/s, F -400N, P ?
• c. P F x V
• c. P -400N (0.4m/s)
• c. P -160 Nm/s or -160 Watts

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Question 8

• A skydiver weighing 170 pounds reaches a constant
  downward velocity of 45m/s. At this velocity the
  upward drag equals the downward weight. What
  power is developed by drag?

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Answer 8

• F 170lbs, V 45m/s, P ?
• Convert pounds to Newtons
• 170lbs X 4.45N/1lb 756.5 N
• P F x V
• P -756.5N (45m/s)
• P -34,042.50 J/s or Watts

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Conclusion