Question

a man is pulling a sled as shown in the diagram below. if he pulls the sled a distance of 2 meters, how much work has he done. (type your numerical answer in the box without the units, rounded to the nearest whole number.) answer: if \\(\vec{a} = (6, -5, 1)\\) and \\(\vec{b} = (-8, -9, 5)\\), answer the questions below: what is \\(\vec{a} \bullet \vec{b}\\)? what is the angle between \\(\vec{a}\\) and \\(\vec{b}\\)? (give your answer to the nearest tenth of a degree)

Explanation:

First Problem (Work Done)

Step1: Recall Work Formula

Work \( W \) is calculated as \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is force, \( d \) is distance, and \( \theta \) is the angle between force and displacement.

Step2: Substitute Values

Given \( F = 11.0 \, \text{N} \), \( d = 2 \, \text{m} \), \( \theta = 29.0^\circ \).
\( \cos(29.0^\circ) \approx 0.8746 \).
\( W = 11.0 \times 2 \times 0.8746 \).

Step3: Calculate

\( 11.0 \times 2 = 22 \); \( 22 \times 0.8746 \approx 19.24 \), rounded to nearest whole number is 19.

Second Problem (Dot Product \( \vec{a} \cdot \vec{b} \))

Step1: Recall Dot Product Formula

For vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \), \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

Step2: Substitute Values

\( \vec{a} = (6, -5, 1) \), \( \vec{b} = (-8, -9, 5) \).
\( \vec{a} \cdot \vec{b} = (6)(-8) + (-5)(-9) + (1)(5) \).

Step3: Calculate

\( 6(-8) = -48 \); \( -5(-9) = 45 \); \( 1(5) = 5 \).
Sum: \( -48 + 45 + 5 = -98 \).

Third Problem (Angle Between \( \vec{a} \) and \( \vec{b} \))

Step1: Recall Angle Formula

\( \cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|} \), where \( \|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \).

Step2: Find Magnitudes

\( \|\vec{a}\| = \sqrt{6^2 + (-5)^2 + 1^2} = \sqrt{36 + 25 + 1} = \sqrt{62} \approx 7.874 \).
\( \|\vec{b}\| = \sqrt{(-8)^2 + (-9)^2 + 5^2} = \sqrt{64 + 81 + 25} = \sqrt{170} \approx 13.038 \).

Step3: Substitute Dot Product and Magnitudes

\( \vec{a} \cdot \vec{b} = -98 \) (from before).
\( \cos(\theta) = \frac{-98}{7.874 \times 13.038} \approx \frac{-98}{102.6} \approx -0.955 \).

Step4: Find Angle

\( \theta = \arccos(-0.955) \approx 162.7^\circ \).

Answer:

19