Each time he does one pushup, Jose, who has a mass of 61kg, raises his center of mass by 25 cm. He completes an impressive set of 150 pushups in 5 minutes, exercising at a steady rate.
Part A
If we assume that lowering his body has no energetic cost, what is his metabolic power during this workout?
Part B
In fact, it costs Jose a certain amount of energy to lower his body-about half of what it costs to raise it. If you include this in your calculation, what is his metabolic power?
Express your answer with the appropriate units.
power = weight*distance / time
Part A:
To calculate the metabolic power during the workout, we need to determine the work done in raising Jose's center of mass during each pushup and then multiply it by the rate at which he completes pushups.
The work done in raising his center of mass can be calculated using the formula:
Work = Force * Distance
In this case, the distance is the height raised by the center of mass, which is 25 cm (0.25 m). The force can be calculated using the formula:
Force = Mass * Acceleration
Acceleration can be calculated using the formula:
Acceleration = (Change in velocity) / (Time)
Since Jose raises his center of mass at a steady rate and we know the distance raised and time taken, we can calculate the acceleration.
Acceleration = 0.25 m / (5 min * 60 s/min) = 0.00833 m/s^2
Plugging these values into the Force formula:
Force = 61 kg * 0.00833 m/s^2 = 0.508 kg·m/s^2
Now we can calculate the work done:
Work = 0.508 kg·m/s^2 * 0.25 m = 0.127 kg·m^2/s^2 (Joules)
Since Jose completes 150 pushups in 5 minutes, the rate at which he completes pushups is:
Rate = 150 pushups / (5 min * 60 s/min) = 0.5 pushups/s
To calculate the metabolic power, we multiply the work done per pushup by the rate at which he completes pushups:
Metabolic Power = Work * Rate = 0.127 kg·m^2/s^2 * 0.5 pushups/s = 0.0635 kg·m^2/s^3 (Watts)
Therefore, Jose's metabolic power during the workout, assuming lowering his body has no energetic cost, is 0.0635 Watts.
Part B:
If we include the cost of lowering his body, it would be half of the work done in raising his center of mass. Therefore, the work done in lowering his body would be:
Work_lowering = 0.5 * 0.127 kg·m^2/s^2 = 0.0635 kg·m^2/s^2 (Joules)
To calculate the total work done (including both raising and lowering his center of mass), we add the work done in raising and lowering:
Total Work = Work + Work_lowering = 0.127 kg·m^2/s^2 + 0.0635 kg·m^2/s^2 = 0.1905 kg·m^2/s^2 (Joules)
Using the same rate of completing pushups as before, we can calculate the metabolic power:
Metabolic Power = Total Work * Rate = 0.1905 kg·m^2/s^2 * 0.5 pushups/s = 0.09525 kg·m^2/s^3 (Watts)
Therefore, Jose's metabolic power during the workout, including the cost of lowering his body, is 0.09525 Watts.
Part A:
To find Jose's metabolic power during the workout, we need to determine the work done in lifting his body during each pushup and then calculate the power.
The work done in raising his body during each pushup can be calculated using the formula:
Work = force x distance
The force required to lift his body can be determined using Newton's second law:
Force = mass x acceleration
The acceleration can be found using the formula:
acceleration = change in velocity / time
The change in velocity can be found using the formula for average velocity:
change in velocity = distance / time
Given:
Mass (m) = 61 kg
Distance raised (d) = 0.25 m
Time (t) = 5 minutes = 300 seconds
First, let's calculate the acceleration:
change in velocity = distance / time
change in velocity = 0.25 m / 300 s = 0.000833 m/s
acceleration = change in velocity / time
acceleration = 0.000833 m/s / 300 s = 2.78 x 10^-6 m/s^2
Next, let's calculate the force:
Force = mass x acceleration
Force = 61 kg x 2.78 x 10^-6 m/s^2 = 1.69 x 10^-4 N
Now, let's calculate the work done:
Work = force x distance
Work = 1.69 x 10^-4 N x 0.25 m = 4.23 x 10^-5 J (Joules)
To find the power, we divide the work by the time:
Power = Work / time
Power = 4.23 x 10^-5 J / 300 s = 1.41 x 10^-7 W (Watts)
Therefore, Jose's metabolic power, assuming lowering his body has no energetic cost, is 1.41 x 10^-7 Watts (W).
Part B:
Now, let's include the energy cost of lowering his body, which is half the energy cost of raising it.
The total work done during each pushup, accounting for both raising and lowering, can be calculated as:
Total work = work raising + work lowering
The work lowering can be calculated as half of the work raising:
Work lowering = 0.5 x work raising
Therefore, the total work can be calculated as:
Total work = work raising + 0.5 x work raising
Total work = 1.5 x work raising
To calculate the power, we divide the total work by the time:
Power = (1.5 x work raising) / time
Using the value of work raising calculated in Part A (4.23 x 10^-5 J), we can substitute it into the equation:
Power = (1.5 x 4.23 x 10^-5 J) / 300 s
Power = 6.34 x 10^-5 J /s (Watts)
Therefore, including the energy cost of lowering his body, Jose's metabolic power is 6.34 x 10^-5 Watts (W).
Part A:
To find Jose's metabolic power during this workout, we need to calculate the work done by raising his center of mass for each push-up.
Work (W) is given by the formula:
W = mgh
where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
In this case, the mass of Jose is 61 kg, and the height raised for each push-up is 25 cm (converted to meters, it is 0.25 m).
So, for each push-up, the work done is:
W = 61 kg * 9.8 m/s^2 * 0.25 m
W = 150.575 J (Joules) (rounded to three decimal places)
Since Jose completes 150 push-ups in 5 minutes, we can calculate the total work using the formula:
Total work = Work per push-up * Number of push-ups
Total work = 150.575 J/push-up * 150 push-ups
Total work = 22,586.25 J
Finally, to find Jose's metabolic power, we divide the total work by the time it took him to complete the workout:
Metabolic power = Total work / Time
Metabolic power = 22,586.25 J / 300 s
Metabolic power = 75.288 W (Watts) (rounded to three decimal places)
Therefore, Jose's metabolic power during this workout, assuming no energetic cost in lowering his body, is approximately 75.288 Watts.
Part B:
If we include the fact that it costs Jose half the energy to lower his body, the total work per push-up changes.
The work done per push-up will now be:
W = (61 kg * 9.8 m/s^2 * 0.25 m) + (0.5 * (61 kg * 9.8 m/s^2 * 0.25 m))
W = 150.575 J + 75.288 J
W = 225.863 J (rounded to three decimal places)
Using the same formula as above, the total work will be:
Total work = 225.863 J/push-up * 150 push-ups
Total work = 33,879.45 J
The metabolic power is then calculated as:
Metabolic power = Total work / Time
Metabolic power = 33,879.45 J / 300 s
Metabolic power = 112.932 W (rounded to three decimal places)
Therefore, considering the energetic cost of lowering his body, Jose's metabolic power during this workout is approximately 112.932 Watts.