Question

Modified Unit 5 Test - Part 2
Complete the following problems, showing all work. Please do not include a commentary of what you are about to do to solve the problem. No credit will be given unless steps are shown. Also, no credit will be given if an Academic Integrity Violation presents itself.

A vector V is given in the coordinate plane. Find the direction angle of vector V, measured counterclockwise from the positive x-axis to the nearest tenth of a degree. (3 pts)











Calculate the work done in Nm to move the block a horizontally distance, d, of 2 meters. Remember to use exact values for the 30 degree angle. (3pts)

Answer 1

### Direction Angle of Vector V

Let \( V = (x, y) \) be the given vector in the coordinate plane.

1. Calculate the magnitude of the vector:
\[
|V| = \sqrt{x^2 + y^2}
\]

2. Find the direction angle \( \theta \) using the tangent function:
\[
\tan(\theta) = \frac{y}{x}
\]

3. Use the inverse tangent function to find the angle:
\[
\theta = \tan^{-1}\left(\frac{y}{x}\right)
\]

4. Adjust the angle based on the quadrant of the vector.

5. Convert to degrees if necessary and round to the nearest tenth.

### Work Done

Let \( F \) be the force applied, and \( d \) be the distance moved.

1. Calculate the horizontal component of the force \( F \):
\[
F_x = F \cdot \cos(30^{\circ}) = F \cdot \frac{\sqrt{3}}{2}
\]

2. Work done \( W \) is given by:
\[
W = F_x \cdot d
\]

3. Substitute \( d = 2 \) meters:
\[
W = F \cdot \frac{\sqrt{3}}{2} \cdot 2 = F \cdot \sqrt{3}
\]

4. If the value of \( F \) is provided, substitute it into the equation to find the exact work done.

The work done will be in Nm (Joules).