College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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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Transcribed Image Text:**Question 9:**
The motor of a boat develops 60 kW when the boat's speed is 15 km/h. With how much force does the water resist the motion of the boat?
![**Physics Problem Example: Calculating Average Power Output**
**Problem Statement:**
11. A weightlifter raises a 70-kg barbell from the floor to a height of 2.2 m in 1.2 s. What was his average power output during the lift?
**Solution:**
To find the weightlifter's average power output, we need to determine the work done and the time over which it was done. Power is given by:
\[ \text{Power (P)} = \frac{\text{Work Done (W)}}{\text{Time (t)}} \]
**Step-by-Step Solution:**
1. **Calculate the Work Done:**
Work done (W) against gravity is given by:
\[ W = m \cdot g \cdot h \]
Where:
- \( m \) = mass of the barbell = 70 kg
- \( g \) = acceleration due to gravity = 9.8 m/s²
- \( h \) = height = 2.2 m
Substituting the values:
\[ W = 70 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2.2 \, \text{m} \]
\[ W = 70 \times 9.8 \times 2.2 \]
\[ W = 1509.6 \, \text{J} (Joules) \]
2. **Calculate the Average Power:**
Using the formula for power:
\[ P = \frac{W}{t} \]
Where:
- \( W \) = 1509.6 J
- \( t \) = time taken = 1.2 s
Substituting the values:
\[ P = \frac{1509.6 \, \text{J}}{1.2 \, \text{s}} \]
\[ P = 1258 \, \text{W} (Watts) \]
Therefore, the weightlifter's average power output during the lift is 1258 W.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1f3a9dd4-0825-4975-b3b1-0ebe0225a05a%2F51ae0f71-6e76-4f16-b0a4-9e56f2b3f5b2%2F7j248x6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics Problem Example: Calculating Average Power Output**
**Problem Statement:**
11. A weightlifter raises a 70-kg barbell from the floor to a height of 2.2 m in 1.2 s. What was his average power output during the lift?
**Solution:**
To find the weightlifter's average power output, we need to determine the work done and the time over which it was done. Power is given by:
\[ \text{Power (P)} = \frac{\text{Work Done (W)}}{\text{Time (t)}} \]
**Step-by-Step Solution:**
1. **Calculate the Work Done:**
Work done (W) against gravity is given by:
\[ W = m \cdot g \cdot h \]
Where:
- \( m \) = mass of the barbell = 70 kg
- \( g \) = acceleration due to gravity = 9.8 m/s²
- \( h \) = height = 2.2 m
Substituting the values:
\[ W = 70 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2.2 \, \text{m} \]
\[ W = 70 \times 9.8 \times 2.2 \]
\[ W = 1509.6 \, \text{J} (Joules) \]
2. **Calculate the Average Power:**
Using the formula for power:
\[ P = \frac{W}{t} \]
Where:
- \( W \) = 1509.6 J
- \( t \) = time taken = 1.2 s
Substituting the values:
\[ P = \frac{1509.6 \, \text{J}}{1.2 \, \text{s}} \]
\[ P = 1258 \, \text{W} (Watts) \]
Therefore, the weightlifter's average power output during the lift is 1258 W.
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