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7 Work Energy: Problem 3
(1 point)
The velocity of a
2.3kg particle changes from
4i-5jm//s to
-6i+3jm//s.
What is its change in kinetic energy?

-893.2
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WeBWork &
1996-2022 | the generated at 11/07/2023 at 06:09pm EST
| theme: math 4 | ww_version: 2.17 I pg_version 2.17 | The WeBWork Project

Previous Problem $\newline$ Problem List $\newline$ Next Problem $\newline$ $7$ Work Energy: Problem $3$ $\newline$ ( $1$ point) $\newline$ The velocity of a $2.3 \mathrm{~kg}$ particle changes from $4 \mathbf{i}-5 \mathbf{j} \mathrm{m} / \mathrm{s}$ to $-6 \mathbf{i}+3 \mathbf{j} \mathrm{m} / \mathrm{s}$ . $\newline$ What is its change in kinetic energy? $\newline$ $-893.2$ $\newline$ Preview My Answers $\newline$ Submit Answers $\newline$ You have attempted this problem $2$ times. $\newline$ Your overall recorded score is $0 \%$ . $\newline$ You have unlimited attempts remaining. $\newline$ Email Instructor $\newline$ WeBWork \& $1996-2022$ | the generated at $11$ / $07$ / $2023$ at $06$ : $09$ pm EST $\newline$ | theme: math $4$ | ww_version: $2$ . $17$ I pg_version $2$ . $17$ | The WeBWork Project

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Transformations of functions

Full solution

Q. Previous Problem

\newline

Problem List

\newline

Next Problem

\newline

7

Work Energy: Problem

3

\newline

(

1

point)

\newline

The velocity of a

2.3 \mathrm{~kg}

particle changes from

4 \mathbf{i}-5 \mathbf{j} \mathrm{m} / \mathrm{s}

to

-6 \mathbf{i}+3 \mathbf{j} \mathrm{m} / \mathrm{s}

.

\newline

What is its change in kinetic energy?

\newline

-893.2

\newline

Preview My Answers

\newline

Submit Answers

\newline

You have attempted this problem

2

times.

\newline

Your overall recorded score is

0 \%

.

\newline

You have unlimited attempts remaining.

\newline

Email Instructor

\newline

WeBWork \&

1996-2022

| the generated at

11

/

07

/

2023

at

06

:

09

pm EST

\newline

| theme: math

4

| ww_version:

2

.

17

I pg_version

2

.

17

| The WeBWork Project

Calculate Initial Kinetic Energy: Calculate the initial kinetic energy using the formula $KE_{\text{initial}} = \frac{1}{2} \cdot m \cdot v_{\text{initial}}^2$ . Initial velocity, $v_{\text{initial}} = 4\mathbf{i} - 5\mathbf{j}$ m/s. So, $v_{\text{initial}}^2 = (4^2 + (-5)^2)$ m $^2$ /s $^2$ . $v_{\text{initial}}^2 = 16 + 25 = 41$ m $^2$ /s $^2$ . $KE_{\text{initial}} = \frac{1}{2} \cdot 2.3\,\text{kg} \cdot 41$ m $^2$ /s $^2$ . $v_{\text{initial}} = 4\mathbf{i} - 5\mathbf{j}$ $1$ . $v_{\text{initial}} = 4\mathbf{i} - 5\mathbf{j}$ $2$ J.
Calculate Final Kinetic Energy: Calculate the final kinetic energy using the formula $KE_{\text{final}} = \frac{1}{2} \cdot m \cdot v_{\text{final}}^2$ . Final velocity, $v_{\text{final}} = -6i + 3j \, \text{m/s}$ . So, $v_{\text{final}}^2 = ((-6)^2 + 3^2) \, \text{m}^2/\text{s}^2$ . $v_{\text{final}}^2 = 36 + 9 = 45 \, \text{m}^2/\text{s}^2$ . $KE_{\text{final}} = \frac{1}{2} \cdot 2.3\,\text{kg} \cdot 45 \, \text{m}^2/\text{s}^2$ . $KE_{\text{final}} = 1.15 \cdot 45$ . $KE_{\text{final}} = 51.75 \, \text{J}$ .
Find Change in Kinetic Energy: Find the change in kinetic energy by subtracting $KE_{\text{initial}}$ from $KE_{\text{final}}$ . $\newline$ Change in $KE = KE_{\text{final}} - KE_{\text{initial}}$ . $\newline$ Change in $KE = 51.75\,\text{J} - 47.15\,\text{J}$ . $\newline$ Change in $KE = 4.6\,\text{J}$ .

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