91Ó°ÊÓ

A machine carries a \(4.0 \mathrm{~kg}\) package from an initial position of \(\vec{d}_{i}=(0.50 \mathrm{~m}) \hat{\mathrm{i}}+(0.75 \mathrm{~m}) \hat{\mathrm{j}}+(0.20 \mathrm{~m}) \hat{\mathrm{k}}\) at \(t=0\) to a final position of \(\vec{d}_{f}=(7.50 \mathrm{~m}) \hat{\mathrm{i}}+(12.0 \mathrm{~m}) \hat{\mathrm{j}}+(7.20 \mathrm{~m}) \hat{\mathrm{k}}\) at \(t=12 \mathrm{~s}\). The constant force applied by the machine on the package is \(\vec{F}=(2.00 \mathrm{~N}) \hat{\mathrm{i}}+(4.00 \mathrm{~N}) \hat{\mathrm{j}}+(6.00 \mathrm{~N}) \hat{\mathrm{k}} .\) For that displacement, find (a) the work done on the package by the machine's force and (b) the average power of the machine's force on the package.

Step by step solution

01

Identify the Displacement Vector

The displacement vector \( \vec{d} \) is the difference between the final position \( \vec{d}_{f} \) and the initial position \( \vec{d}_{i} \). Calculate it as follows:\[ \vec{d} = \vec{d}_{f} - \vec{d}_{i} = [(7.50 - 0.50) \hat{\mathrm{i}} + (12.0 - 0.75) \hat{\mathrm{j}} + (7.20 - 0.20) \hat{\mathrm{k}}] = (7.00 \mathrm{~m}) \hat{\mathrm{i}} + (11.25 \mathrm{~m}) \hat{\mathrm{j}} + (7.00 \mathrm{~m}) \hat{\mathrm{k}}. \]

02

Calculate the Work Done by the Force

The work \( W \) done is given by the dot product of force \( \vec{F} \) and displacement \( \vec{d} \).\[ W = \vec{F} \cdot \vec{d} = (2.00 \hat{\mathrm{i}} + 4.00 \hat{\mathrm{j}} + 6.00 \hat{\mathrm{k}}) \cdot (7.00 \hat{\mathrm{i}} + 11.25 \hat{\mathrm{j}} + 7.00 \hat{\mathrm{k}}) \]Calculate the dot product:\[ W = (2.00 \times 7.00) + (4.00 \times 11.25) + (6.00 \times 7.00) \]\[ W = 14.00 + 45.00 + 42.00 = 101.00 \mathrm{~J}. \]

03

Determine the Average Power

The average power \( P_{avg} \) is the work done divided by the time interval over which the work is performed. Here, the time interval \( \Delta t = 12 \mathrm{~s} \).\[ P_{avg} = \frac{W}{\Delta t} = \frac{101.00 \mathrm{~J}}{12 \mathrm{~s}} \approx 8.42 \mathrm{~W}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Displacement Vector Calculation

In physics, the displacement vector is essential for describing the change in position of an object over time. It helps in understanding how far and in what direction the object has moved.
To find the displacement vector, we need to calculate the difference between the final and initial position vectors. This involves vector subtraction.

• The initial position vector, \( \vec{d}_{i} \), and the final position vector, \( \vec{d}_{f} \), are given in components along the \( \hat{\mathrm{i}} \), \( \hat{\mathrm{j}} \), and \( \hat{\mathrm{k}} \) directions.
• By subtracting the components of \( \vec{d}_{i} \) from \( \vec{d}_{f} \), we obtain the displacement vector \( \vec{d} = (7.00 \, \mathrm{m}) \hat{\mathrm{i}} + (11.25 \, \mathrm{m}) \hat{\mathrm{j}} + (7.00 \, \mathrm{m}) \hat{\mathrm{k}} \).

The displacement vector provides not only the magnitude of movement but also the specific path taken along each spatial direction.

Dot Product in Physics

The dot product is a fundamental operation in physics, often used for finding work done by a force acting along a displacement.
It is an algebraic calculation that multiplies two vectors to produce a scalar quantity.
In this problem, the dot product of the force vector \( \vec{F} \) and displacement vector \( \vec{d} \) gives the work \( W \) done by the force on the package.

• To perform a dot product, multiply the corresponding components of the vectors: \( (2.00 \, \hat{\mathrm{i}} + 4.00 \, \hat{\mathrm{j}} + 6.00 \, \hat{\mathrm{k}}) \cdot (7.00 \, \hat{\mathrm{i}} + 11.25 \, \hat{\mathrm{j}} + 7.00 \, \hat{\mathrm{k}}) \).
• The resulting scalar is the sum of these products: \( 2.00 \times 7.00 + 4.00 \times 11.25 + 6.00 \times 7.00 = 101.00 \mathrm{~J} \).

This scalar value indicates the total amount of energy transferred by the force to move the object.

Average Power Formula

Power in physics is the rate at which work is done or energy is transferred over time.
The concept of average power is crucial when work is done over a period rather than instantaneously.
To determine the average power \( P_{avg} \) delivered by a force, divide the total work done \( W \) by the time period \( \Delta t \) over which the work is carried out.

• In this context, the work done \( W \) is \( 101.00 \, \mathrm{J} \), and the time interval is \( 12 \, \mathrm{s} \).
• The average power is calculated as \( P_{avg} = \frac{101.00 \, \mathrm{J}}{12 \, \mathrm{s}} \approx 8.42 \, \mathrm{~W} \).

This calculation is vital for understanding the machine's efficiency in doing work within the given timeframe.