Chapter 6: Problem 133
(a) At what rate must you lift a \(3.6-\mathrm{kg}\) container of milk (1 gallon) if the power output of your arm is to be \(22 \mathrm{~W}\) ? (b) How long does it take to lift the milk container through a distance of \(1.0 \mathrm{~m}\) at this rate?
Short Answer
Expert verified
(a) Lift at 0.624 m/s. (b) Takes 1.60 seconds.
Step by step solution
01
Understanding Power and Force
Power (P) is defined as the rate at which work is done. The formula for power is given by \( P = \frac{W}{t} \), where \( W \) is work and \( t \) is time. Work done is equal to the force applied over the distance, which can be expressed as \( W = F \, d \). Since we are lifting vertically, force \( F = m \, g \), where \( m \) is mass and \( g \) is gravitational acceleration \( 9.8 \text{ m/s}^2 \).
02
Calculate the Lifting Force Required
First, calculate the force needed to lift the container using the formula: \( F = m \cdot g \). Given \( m = 3.6 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \), we compute \( F = 3.6 \cdot 9.8 = 35.28 \text{ N} \).
03
Relate Power to Force and Velocity
Power can also be expressed in terms of force and velocity ( v) as \( P = F \cdot v \). We are given \( P = 22 \text{ W} \) and have calculated \( F = 35.28 \text{ N} \). Substitute these into the power equation: \( 22 = 35.28 \cdot v \).
04
Calculate Velocity
Solve for velocity (v): \( v = \frac{P}{F} = \frac{22}{35.28} \approx 0.624 \text{ m/s} \). This is the rate at which you must lift the milk container.
05
Calculate Time to Lift Container
To find out how long it takes to lift the container, use \( t = \frac{d}{v} \), where \( d = 1.0 \text{ m} \) and \( v = 0.624 \text{ m/s} \). Substitute the known values into the equation to find \( t \): \( t = \frac{1.0}{0.624} \approx 1.60 \text{ seconds} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power
Power is a fundamental concept in physics. It refers to the rate at which work is accomplished or energy is transferred. Think of it as how "fast" work is being done. The unit of power is the watt (W), where 1 watt is equivalent to 1 joule of work done per second.
Consider the formula for power:
Consider the formula for power:
- \( P = \frac{W}{t} \), where:
- \( P \) is power,
- \( W \) is work,
- \( t \) is time.
Work
Work is about moving something using a force. It’s a straightforward concept but vital in physics. When we do work, energy is transferred from one place to another or from one form to another. The basic formula for work is:
- \( W = F \cdot d \), where:
- \( W \) is work,
- \( F \) is the force applied,
- \( d \) is the distance moved in the direction of the force.
Gravitational Force
Gravitational force is the force of attraction between two masses. On Earth, this force gives weight to objects and tries to pull everything toward the center of the planet. The formula for calculating gravitational force is:
- \( F = m \, g \), where:
- \( F \) is the gravitational force,
- \( m \) is mass,
- \( g \) is the gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \) on Earth).
Velocity
Velocity is a vector quantity that refers to "how fast" something is moving and in "what direction". It’s different from speed because it also considers direction. In practical terms, knowing velocity means you understand how quickly and where something will end up.
The formula that links power, force, and velocity is:
The formula that links power, force, and velocity is:
- \( P = F \cdot v \), where:
- \( P \) is power,
- \( F \) is force,
- \( v \) is velocity.
Time Calculation
Time calculation is crucial to determine "how long" a task takes. When you have a constant rate of doing work, like consistently lifting an object, computing time helps in planning and optimizing activities. The formula to determine time when velocity and distance are known is:
- \( t = \frac{d}{v} \), where:
- \( t \) is time,
- \( d \) is distance,
- \( v \) is velocity.