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After the warm-up the participant performs the WAnT. You don't have equipment to automatically track power output, therefore, you count revolutions for each 5 second period of the test. The results are as follows:

Time(sec) 5sec_rpm cumulative
0-5 10 10
5-10 7 17
10-15 9 26
15-20 7 33
20-25 7 40
25-30 6 46

The power output can now be computed for each 5 second period using the following equation:

kg of resistance x number of revolutions x 11.765= power (watts)

Required:
What is the power output for each 5 second period?

User Jpsimons
by
7.6k points

1 Answer

2 votes

Answer: Missing detail ( weight of resistance = 6.69 Kg )

answer :

  • 787.0785 watts
  • 550.95 watts
  • 708.37 watts
  • 550.95 watts
  • 550.95 watts
  • 472.25 watts

Explanation:

Applying the equation for computing the power output

= Kg of resistance * number of revolutions * 11.765 ------- ( 1 )

Calculate the power output for each 5 second period

weight ( Kg of resistance ) = 6.69 Kg

Equation 1 becomes :

6.69 * number of revolutions * 11.765 ---- ( 2 )

Next :Apply equation (2) to every 5 seconds interval as represented in the table

For Time 0 - 5

Number of revolutions ( rpm ) = 10

power output = 6.69 * 10 * 11.765 = 787.0785 watts

cumulative rpm = 10 + 0

Cumulative Power = power output + previous power output = 787. 0785watt

For Time ( 5 - 10 )

Number of revolutions ( rpm ) = 7

Power output = 6.69 * 7 * 11.765 = 550.95 watts

Cumulative rpm = rpm + previous rpm = 7 + 10 = 17

Cumulative power = power output + previous power output

= 550.95 + 787.0785 = 1338.0285 watts

Power output for ( 10 - 15 ) = 6.69 * 9 * 11.765 = 708.37 watts

Power output for ( 15 - 20 ) = 6.69 * 7 * 11.765 = 550.95 watts

Power output for ( 20 - 25 ) = 6.69 * 7 * 11.765 = 550.95 watts

Power output for ( 25 - 30 ) = 6.69 * 6 * 11.765 = 472.25 watts

User Ooba Elda
by
8.1k points

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