• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

### Select your language

Suggested languages for you:

Americas

Europe

11P

Expert-verified
Found in: Page 171

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A 12.0 N force with a fixed orientation does work on a particle as the particle moves through the three-dimensional displacement $\stackrel{\mathbf{\to }}{\mathbf{d}}{\mathbf{=}}\left(2.00\stackrel{^}{i}-4.00\stackrel{^}{j}+30.0\stackrel{^}{k}\right)$. What is the angle between the force and the displacement if the change in the particle’s kinetic energy is (a) +30.0 J and (b) -30.0 J ?

1. The angle between the force and the displacement is $\mathrm{\varphi }=62.{3}^{°}$.
2. The angle between the force and the displacement is $\mathrm{\varphi }={118}^{°}$.
See the step by step solution

## Step 1: Given data:

The force, $\mathrm{F}=12\mathrm{N}$

The displacement, $\stackrel{\to }{\mathrm{d}}=\left(2.00\stackrel{^}{\mathrm{i}}-4.00\stackrel{^}{\mathrm{j}}+30.0\stackrel{^}{\mathrm{k}}\right)$

## Step 2: Understanding the concept:

The work-energy theorem states that the net work done by forces on an object is equal to the change in its kinetic energy.

Using the work-kinetic energy theorem, you have

$\begin{array}{rcl}{\mathbf{∆}}{\mathbf{K}}& {\mathbf{=}}& {\mathbf{W}}\\ & {\mathbf{=}}& \stackrel{\mathbf{\to }}{\mathbf{F}}{\mathbf{.}}\stackrel{\mathbf{\to }}{\mathbf{d}}\\ & {\mathbf{=}}& {\mathbf{Fdcos\varphi }}\end{array}$

## Step 3: The magnitude of the displacement:

Calculate the magnitude of the displacement as below.

$\begin{array}{rcl}\mathrm{d}& =& \sqrt{{\left(2.00\mathrm{m}\right)}^{2}+{\left(-4.00\mathrm{m}\right)}^{2}+{\left(3.00\mathrm{m}\right)}^{2}}\\ & =& \sqrt{\left(4+16+9\right)}\mathrm{m}\\ & =& \sqrt{29}\mathrm{m}\\ & =& 5.39\mathrm{m}\end{array}$

## Step 4: (a) The angle between the force and the displacement if the kinetic energy is +30.0 J :

The angle between the force and the displacement if the change in the particle’s kinetic energy is,

$∆\mathrm{K}=+30.0\mathrm{J}$

Therefore, the angle is,

$\begin{array}{rcl}\mathrm{\varphi }& =& {\mathrm{cos}}^{-1}\left(\frac{∆\mathrm{K}}{\mathrm{Fd}}\right)\\ & =& {\mathrm{cos}}^{-1}\left(\frac{30.0\mathrm{J}}{\left(12.0\mathrm{N}\right)\left(5.39\mathrm{m}\right)}\right)\\ & =& {\mathrm{cos}}^{-1}\left(0.464\right)\\ & =& 62.{3}^{°}\end{array}$

Hence, the angle between the force and the displacement is $62.{3}^{°}$.

## Step 5: (b) The angle between the force and the displacement if the kinetic energy is -30.0 J :

The angle between the force and the displacement if the change in the particle’s kinetic energy is,

$∆\mathrm{K}=-30.0\mathrm{J}$

Therefore, the angle is,

$\begin{array}{rcl}\mathrm{\varphi }& =& {\mathrm{cos}}^{-1}\left(\frac{∆\mathrm{K}}{\mathrm{Fd}}\right)\\ & =& {\mathrm{cos}}^{-1}\left(\frac{-30.0\mathrm{J}}{\left(12.0\mathrm{N}\right)\left(5.39\mathrm{m}\right)}\right)\\ & =& {\mathrm{cos}}^{-1}\left(-0.464\right)\\ & =& 117.{64}^{°}\\ \mathrm{\varphi }& \approx & {118}^{°}\end{array}$

Hence, the angle between the force and the displacement is ${118}^{°}$.