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A force \(\vec{F}=(2.00 \hat{\mathrm{i}}+9.00 \hat{\mathrm{j}}+5.30 \hat{\mathrm{k}}) \mathrm{N}\) acts on a \(2.90 \mathrm{~kg}\) object that moves in time interval \(2.10 \mathrm{~s}\) from an initial position \(\vec{r}_{1}=(2.70 \hat{\mathrm{i}}-2.90 \hat{\mathrm{j}}+5.50 \hat{\mathrm{k}}) \mathrm{m}\) to a final position \(\vec{r}_{2}=\) \((-4.10 \hat{\mathrm{i}}+3.30 \hat{\mathrm{j}}+5.40 \hat{\mathrm{k}}) \mathrm{m} .\) Find (a) the work done on the object by the force in that time interval, (b) the average power due to the force during that time interval, and (c) the angle between vectors \(\vec{r}_{1}\) and \(\vec{r}_{2}\)

Short Answer

Expert verified
(a) 41.67 J, (b) 19.84 W, (c) 79.76°

Step by step solution

01

Calculate the Displacement Vector

The displacement vector \( \vec{d} \) is calculated by subtracting the initial position vector \( \vec{r}_1 \) from the final position vector \( \vec{r}_2 \). This is done as follows:\[ \vec{d} = \vec{r}_2 - \vec{r}_1 = \left((-4.10 \hat{\mathrm{i}}+3.30 \hat{\mathrm{j}}+5.40 \hat{\mathrm{k}}) - (2.70 \hat{\mathrm{i}}-2.90 \hat{\mathrm{j}}+5.50 \hat{\mathrm{k}}) \right) \]\[ \vec{d} = (-4.10 - 2.70) \hat{\mathrm{i}} + (3.30 + 2.90) \hat{\mathrm{j}} + (5.40 - 5.50) \hat{\mathrm{k}} \]\[ \vec{d} = (-6.80 \hat{\mathrm{i}} + 6.20 \hat{\mathrm{j}} - 0.10 \hat{\mathrm{k}}) \text{ m} \]
02

Calculate the Work Done

Work done \( W \) by a force is given by the dot product of the force vector \( \vec{F} \) and the displacement vector \( \vec{d} \):\[ W = \vec{F} \cdot \vec{d} = (2.00 \hat{\mathrm{i}}+9.00 \hat{\mathrm{j}}+5.30 \hat{\mathrm{k}}) \cdot (-6.80 \hat{\mathrm{i}} + 6.20 \hat{\mathrm{j}} - 0.10 \hat{\mathrm{k}}) \]Calculating the dot product:\[ W = (2.00)(-6.80) + (9.00)(6.20) + (5.30)(-0.10) \]\[ W = -13.60 + 55.80 - 0.53 \]\[ W = 41.67 \text{ J} \]
03

Calculate the Average Power

Average power \( P_{avg} \) is defined as the work done divided by the time interval \( \Delta t \):\[ P_{avg} = \frac{W}{\Delta t} = \frac{41.67}{2.10} \]\[ P_{avg} \approx 19.84 \text{ W} \]
04

Calculate the Magnitudes of the Vectors

To find the angle between \( \vec{r}_1 \) and \( \vec{r}_2 \), we first calculate the magnitudes of these vectors:Magnitude of \( \vec{r}_1 \):\[ |\vec{r}_1| = \sqrt{(2.70)^2 + (-2.90)^2 + (5.50)^2} \]\[ |\vec{r}_1| = \sqrt{7.29 + 8.41 + 30.25} = \sqrt{45.95} \]\[ |\vec{r}_1| \approx 6.78 \text{ m} \]Magnitude of \( \vec{r}_2 \):\[ |\vec{r}_2| = \sqrt{(-4.10)^2 + (3.30)^2 + (5.40)^2} \]\[ |\vec{r}_2| = \sqrt{16.81 + 10.89 + 29.16} = \sqrt{56.86} \]\[ |\vec{r}_2| \approx 7.54 \text{ m} \]
05

Calculate the Dot Product of \( \vec{r}_1 \) and \( \vec{r}_2 \)

Calculate the dot product \( \vec{r}_1 \cdot \vec{r}_2 \):\[ \vec{r}_1 \cdot \vec{r}_2 = (2.70)(-4.10) + (-2.90)(3.30) + (5.50)(5.40) \]\[ \vec{r}_1 \cdot \vec{r}_2 = -11.07 - 9.57 + 29.70 \]\[ \vec{r}_1 \cdot \vec{r}_2 = 9.06 \text{ m}^2 \]
06

Calculate the Angle Between Vectors

The angle \( \theta \) between two vectors is calculated using the formula:\[ \cos(\theta) = \frac{\vec{r}_1 \cdot \vec{r}_2}{|\vec{r}_1||\vec{r}_2|} \]Substitute the computed values:\[ \cos(\theta) = \frac{9.06}{6.78 \times 7.54} \]\[ \cos(\theta) = \frac{9.06}{51.10} \approx 0.177 \]\[ \theta = \cos^{-1}(0.177) \approx 79.76^\circ \]

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Key Concepts

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

Vector Displacement
In physics, the displacement vector characterizes the change in position of an object. It provides both the direction and the magnitude of the movement between two points. This is different from distance, which only describes how much ground an object has covered, regardless of its direction.

To find the vector displacement, you subtract the initial position vector from the final position vector. Consider a scenario where an object moves from an initial position \( \vec{r}_1 \) to a final position \( \vec{r}_2 \). The displacement \( \vec{d} \) is given by the formula:
\[ \vec{d} = \vec{r}_2 - \vec{r}_1 \]

It's important to remember:
  • Direction Matters: Displacement considers the direction of the movement, making it a vector quantity.
  • Initial and Final Positions: Ensure correct vector subtraction by aligning all vector components (i.e., \( \, \hat{\mathrm{i}} \, , \, \hat{\mathrm{j}} \, , \, \hat{\mathrm{k}} \, \)).
  • Units: Displacement is measured in meters (m).

In our case, the displacement vector is calculated as \( \vec{d} = (-6.80 \, \hat{\mathrm{i}} + 6.20 \, \hat{\mathrm{j}} - 0.10 \, \hat{\mathrm{k}}) \, \text{m} \). This indicates that the object moved predominantly in the negative x-direction (\( \hat{\mathrm{i}} \)), positively in the y-direction (\( \hat{\mathrm{j}} \)), and slightly moved down in the z-direction (\( \hat{\mathrm{k}} \)).
Dot Product
The dot product is a fundamental mathematical operation used to calculate the amount of one vector that extends in the direction of another. It's particularly useful in physics to determine the work done by a force acting on an object during its displacement. The dot product is calculated by multiplying the corresponding components of two vectors, and summing these products.

The typical formula for finding the dot product \( \vec{A} \cdot \vec{B} \) is:
\[ \vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z \]

The dot product can define the work \( W \) done by a force \( \vec{F} \) along a certain displacement \( \vec{d} \):
\[ W = \vec{F} \cdot \vec{d} \]

Key aspects of the dot product include:
  • No Direction Content: The dot product is a scalar. It provides magnitude but no directional information.
  • Non-Perpendicular Contribution: Only the portion of vectors that are in the same direction contributes to the dot product.

In this problem, \( W = 41.67 \, \text{J} \), which reflects the amount of work performed by the force on the object.
Average Power
Average power is a representation of the work done by a force over time. It's an important concept in understanding how energy is transferred per unit of time. By determining average power, we understand not only the work done but also the speed of this energy conversion.

The formula for average power \( P_{\text{avg}} \) is:
\[ P_{\text{avg}} = \frac{W}{\Delta t} \]

Where:
  • \( W \) represents the work done, measured in Joules (J).
  • \( \Delta t \) is the time interval during which the work was done, measured in seconds (s).

This formula emphasizes:
  • Time Dependency: Power depends on both the work done and the time taken; greater power means faster energy transfer.
  • Units: Power is measured in watts (W), with 1 W equating to 1 Joule per second.

Following through on our example, we calculate the average power to be around \( 19.84 \) watts, signifying that energy was transferred at a rate of roughly 19.84 Joules per second over the given time interval.

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Most popular questions from this chapter

A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) \(2.5 \mathrm{~N} / \mathrm{cm}\) (Fig. \(7-46)\). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

A force of \(5.0 \mathrm{~N}\) acts on a \(15 \mathrm{~kg}\) body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

In the block-spring arrangement of Fig. \(7-10,\) the block's mass is \(4.00 \mathrm{~kg}\) and the spring constant is \(500 \mathrm{~N} / \mathrm{m} .\) The block is released from position \(x_{i}=0.300 \mathrm{~m} .\) What are \((\mathrm{a})\) the block's speed at \(x=0,\) (b) the work done by the spring when the block reaches \(x=0,\) (c) the instantaneous power due to the spring at the release point \(x_{i}\), (d) the instantaneous power at \(x=0,\) and \((\mathrm{e})\) the block's position when the power is maximum?

A particle moves along a straight path through displacement \(\vec{d}=(8 \mathrm{~m}) \hat{\mathrm{i}}+c \hat{\mathrm{j}}\) while force \(\vec{F}=(2 \mathrm{~N}) \hat{\mathrm{i}}-(4 \mathrm{~N}) \hat{\mathrm{j}}\) acts on it. (Other forces also act on the particle.) What is the value of \(c\) if the work done by \(\vec{F}\) on the particle is (a) zero, (b) positive, and (c) negative?

If a car of mass \(1200 \mathrm{~kg}\) is moving along a highway at \(120 \mathrm{~km} / \mathrm{h}\) what is the car's kinetic energy as determined by someone standing alongside the highway?

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