(a) What is the average useful power output (in W) of a person who does 6.80 x 106 J of useful work in 8.40 h? W (b) Working at this rate, how long (in s) will it take this person to lift 1700 kg of bricks 1.40 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)
(a) What is the average useful power output (in W) of a person who does 6.80 x 106 J of useful work in 8.40 h? W (b) Working at this rate, how long (in s) will it take this person to lift 1700 kg of bricks 1.40 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Educational Task: Calculating Power Output and Time for Lifting
#### (a) Calculation of Average Useful Power Output
**Problem Statement:**
What is the average useful power output (in watts, W) of a person who performs \(6.80 \times 10^6\) joules (J) of useful work in 8.40 hours?
**Solution:**
To find the average power output, use the formula:
\[ \text{Power} = \frac{\text{Work}}{\text{Time}} \]
- Work = \(6.80 \times 10^6\) J
- Time = 8.40 hours
*Note:* Convert the time from hours to seconds (since power in watts is joules per second).
\[ 8.40 \, \text{hours} = 8.40 \times 3600 \, \text{seconds} \]
Fill in the answer box with the calculated power in watts (W).
#### (b) Calculation of Time Required for Lifting
**Problem Statement:**
Working at this rate, how long (in seconds, s) will it take this person to lift 1700 kg of bricks 1.40 meters to a platform? (Work done to lift their body can be omitted because it is not considered useful output here.)
**Solution:**
To find the time required, use the formula:
\[ \text{Time} = \frac{\text{Work}}{\text{Power}} \]
First, calculate the work required to lift the bricks:
\[ \text{Work} = \text{mass} \times \text{gravity} \times \text{height} \]
- Mass = 1700 kg
- Gravity = 9.8 m/s\(^2\)
- Height = 1.40 m
\[ \text{Work} = 1700 \times 9.8 \times 1.40 \, \text{J} \]
Use the previously calculated power from part (a) to find the time.
Fill in the answer box with the calculated time in seconds (s).
### Note:
These calculations help understand the relationship between work, power, and time in physical processes. Ensure all units are consistent while calculating.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fdc4c1b-93d7-475e-a53d-95a124c7e126%2Fde6a7825-e984-49ff-96d0-4ceb11652076%2Fzm5hxei_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Task: Calculating Power Output and Time for Lifting
#### (a) Calculation of Average Useful Power Output
**Problem Statement:**
What is the average useful power output (in watts, W) of a person who performs \(6.80 \times 10^6\) joules (J) of useful work in 8.40 hours?
**Solution:**
To find the average power output, use the formula:
\[ \text{Power} = \frac{\text{Work}}{\text{Time}} \]
- Work = \(6.80 \times 10^6\) J
- Time = 8.40 hours
*Note:* Convert the time from hours to seconds (since power in watts is joules per second).
\[ 8.40 \, \text{hours} = 8.40 \times 3600 \, \text{seconds} \]
Fill in the answer box with the calculated power in watts (W).
#### (b) Calculation of Time Required for Lifting
**Problem Statement:**
Working at this rate, how long (in seconds, s) will it take this person to lift 1700 kg of bricks 1.40 meters to a platform? (Work done to lift their body can be omitted because it is not considered useful output here.)
**Solution:**
To find the time required, use the formula:
\[ \text{Time} = \frac{\text{Work}}{\text{Power}} \]
First, calculate the work required to lift the bricks:
\[ \text{Work} = \text{mass} \times \text{gravity} \times \text{height} \]
- Mass = 1700 kg
- Gravity = 9.8 m/s\(^2\)
- Height = 1.40 m
\[ \text{Work} = 1700 \times 9.8 \times 1.40 \, \text{J} \]
Use the previously calculated power from part (a) to find the time.
Fill in the answer box with the calculated time in seconds (s).
### Note:
These calculations help understand the relationship between work, power, and time in physical processes. Ensure all units are consistent while calculating.
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