A digital video camera with a shutter speed of 1/500 seconds was used to capture

Question

A digital video camera with a shutter speed of 1/500 seconds was used to capture two frames (images) of an individual doing a squat exercise. This video was then processed in a data analysis computer program. With the given information, please determine the work performed and power generated in lifting a specific weight from position 1 to position 2. (specific weight is a 45 pound barbell during a squat exercise)

Position

Phase

Time (seconds)

X coordinate (meters)

Y coordinate (meters)

1

Squat (lowest point)

8.39

1.176

0.739

2

Squat (highest point)

9.66

1.092

1.317

 ***Please provide all formulas/units used***

***let me know if theres any other information that I need to provide....I understand how to get power and work...but I don't know if I'm supposed to use the shutter speed of the camera or not in my calculations***

Position

Phase

Time (seconds)

X coordinate (meters)

Y coordinate (meters)

1

Squat (lowest point)

8.39

1.176

0.739

2

Squat (highest point)

9.66

1.092

1.317

Explanation / Answer

First off, as you already know the time elapsed between the two moments, shutter speed will only matter in calculating margin of error. Because work equates to force applied over a distance (W = Fd), you must decipher from this data what distance was crossed and what force did the crossing. In this case, the barbell traveled in both the x and y directions, requiring an analysis of the two vectors involved. In y, the squatter acted against gravity, the force being a product of gravitational acceleration and mass (F = ma), which, for the sake of loving SI units furiously, I have converted pounds to kilograms (45 lbs * .4539 kg/lb = 20.41166 Kg). Multiplied by earth's gravity's acceleration, 9.8 m/s^2, the force results: 200.03424 N Multiply by the distance: W = 200.03424 N * (1.317 m - 0.739 m) W = 115.6198 Joules We've worked out the work! Now for your x coordinate movement, the actual force applied would vary based on the acceleration the barbell underwent (incalculable from the data given) because if the barbell chugged along with a constant velocity, no acceleration would act upon it, and according to F = ma, if a = 0 then F = 0, but conversely, it migh have vibrated infinitely quickly between the two positions at which it began and finished, providing an infinite acceleration. For the purpose of the experiment, I would use the small variance in x and the lack of horizontal shiver of the weightlifter (a subjective observation, but a valid scientific one nonetheless) to argue that the x direction work and power are negligible. Power follows easily. The equation: Power = work/time Throw in that elapsed time, a teensy 1.27 seconds: P = 115.6198 J / 1.26 s P = 91.8 watts The moment we've all been WATTing for! Anyway, that was quite the neat application of the O.G. himself, Newton's, system of equations! Best of luck in your kinesthesiology endeavors (presumably) and don't hesitate to ask for clarification!

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