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Found in: Page 150

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# Find the scalar product of the vectors in Figure P7.10.

The scalar product of the two vectors is 5.33 J.

See the step by step solution

## Step 1: Given data

Magnitude of the first vector

A =32.8 N

Magnitude of the second vector

$B=17.3\mathrm{cm}\phantom{\rule{0ex}{0ex}}=17.3·\left(1\mathrm{cm}×\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\phantom{\rule{0ex}{0ex}}=0.173\mathrm{m}$

Large angle made by the first vector and y axis = ${118}^{\circ }$

Large angle made by the second vector and x axis = ${132}^{\circ }$

## Step 2: Dot product of two vectors

The dot product of two vectors $\stackrel{\mathbf{\to }}{\mathbf{A}}$ and $\stackrel{\mathbf{\to }}{\mathbf{B}}$ is

$\stackrel{\mathbf{\to }}{\mathbf{A}}{\mathbf{·}}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{=}}{\mathbf{ABcos\theta }}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}\left(I\right)$

## Step 3: Determining the dot product of the given vectors

Large angle between the positive y axis and the positive x axis = ${270}^{°}$

Thus, angle between the two vectors

$\theta ={270}^{\circ }-{118}^{\circ }-{132}^{\circ }\phantom{\rule{0ex}{0ex}}={20}^{\circ }$

From equation (I), the dot product of the two vectors is

$\stackrel{\to }{A}·\stackrel{\to }{B}=32.8\mathrm{N}×0.173\mathrm{m}×\mathrm{cos}{20}^{\circ }\phantom{\rule{0ex}{0ex}}=5.33·\left(1\mathrm{N}·\mathrm{m}×\frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right)\phantom{\rule{0ex}{0ex}}=5.33\mathrm{J}$

Thus the required dot product is 5.33 J.