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(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:
  • \( P = \frac{W}{t} \), where:
    • \( P \) is power,
    • \( W \) is work,
    • \( t \) is time.
In our exercise, the power output is given as 22 watts. This number tells us how quickly energy is used to lift the container of milk. If power supply is too high or low, either you’ll tire quickly or the task won't be completed in time. Understanding power helps in engineering tasks or even in understanding everyday tasks like lifting objects.
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.
In a vertical lift, the force needed is the gravitational force. So, lifting a gallon of milk involves doing work against gravity to elevate it. Calculating work helps us understand how much energy is used or needed – crucial for both small and large engineering feats.
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).
With a container of milk having a mass of 3.6 kg, the gravitational force is 35.28 N (newtons). Understanding this concept is important in determining how much force is needed to lift objects against Earth's gravity, helping comprehend weights, balances, and the effort needed in different scenarios.
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:
  • \( P = F \cdot v \), where:
    • \( P \) is power,
    • \( F \) is force,
    • \( v \) is velocity.
Here, calculating the necessary velocity to lift the milk container involves rearranging the formula to \( v = \frac{P}{F} \). By inserting the given values, we get approximately 0.624 m/s. This velocity ensures the power supply precisely matches the work needed to continually lift the container. Understanding velocity is fundamental in scenarios ranging from vehicle speeds, race dynamics, to simply moving objects effectively.
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.
In the exercise, knowing the lifting distance is 1 meter and the velocity is 0.624 m/s, we find it takes about 1.6 seconds to lift the milk container. Accurately computing time allows better scheduling and understanding of both productive and inefficient periods in various processes and activities.

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