Lesson 1 : Basic Terminology And Concepts

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9. A tired squirrel (mass of approximately 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm in order to do a mere 0.50 Joule of work. If the tired squirrel does all this work in 2 seconds, then determine its power.

0.25 Watts

0.75 Watts

1.00 Watts

None of the above

The tired squirrel does 0.50 Joule of work in 2.0 seconds. The power rating of this squirrel is found by

P = W / t = (0.50 J) / (2.0 s) = 0.25 Watts

Explanation

The tired squirrel does 0.50 Joule of work in 2.0 seconds. The power rating of this squirrel is found by

P = W / t = (0.50 J) / (2.0 s) = 0.25 Watts

10. When doing a chin-up, a physics student lifts her 42.0-kg body a distance of 0.25 meters in 2 seconds. What is the power delivered by the student's biceps?

51.5 Watts

75.0 Watts

25.0 Watts

None of the above.

To raise her body upward at a constant speed, the student must apply a force which is equal to her weight (m•g). The work done to lift her body is

W = F * d = (411.6 N) * (0.250 m)
W = 102.9 J

The power is the work/time ratio which is (102.9 J) / (2 seconds) = 51.5 Watts (rounded)

Explanation

To raise her body upward at a constant speed, the student must apply a force which is equal to her weight (m•g). The work done to lift her body is

W = F * d = (411.6 N) * (0.250 m)
W = 102.9 J

The power is the work/time ratio which is (102.9 J) / (2 seconds) = 51.5 Watts (rounded)

11. An escalator is used to move 20 passengers every minute from the first floor of a department store to the second. The second floor is located 5.20 meters above the first floor. The average passenger's mass is 54.9 kg. Determine the power requirement of the escalator in order to move this number of passengers in this amount of time.

1000 W

933 W

879 W

None of the above.

A good strategy would involve determining the work required to elevate one average passenger. Then multiply this value by 20 to determine the total work for elevating 20 passengers. Finally, the power can be determined by dividing this total work value by the time required to do the work. The solution goes as follows:

W1 passenger = F • d • cos(0 deg)
W1 passenger = (54.9 kg • 9.8 m/s2) • 5.20 m = 2798 J (rounded)

W20 passengers = 55954 J (rounded)

P = W20 passengers / time = (55954 J) / (60 s)

P = 933 W

Explanation

A good strategy would involve determining the work required to elevate one average passenger. Then multiply this value by 20 to determine the total work for elevating 20 passengers. Finally, the power can be determined by dividing this total work value by the time required to do the work. The solution goes as follows:

W1 passenger = F • d • cos(0 deg)
W1 passenger = (54.9 kg • 9.8 m/s2) • 5.20 m = 2798 J (rounded)

W20 passengers = 55954 J (rounded)

P = W20 passengers / time = (55954 J) / (60 s)

P = 933 W

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