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11P

Expert-verifiedFound in: Page 171

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 $\overrightarrow{\mathbf{d}}{\mathbf{=}}{\left(2.00\hat{i}-4.00\hat{j}+30.0\hat{k}\right)}$. 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 ?**

- The angle between the force and the displacement is $\mathrm{\varphi}=62.{3}^{\xb0}$.
- The angle between the force and the displacement is $\mathrm{\varphi}={118}^{\xb0}$.

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

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

**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**

**$\begin{array}{rcl}{\mathbf{\u2206}}{\mathbf{K}}& {\mathbf{=}}& {\mathbf{W}}\\ & {\mathbf{=}}& \overrightarrow{\mathbf{F}}{\mathbf{.}}\overrightarrow{\mathbf{d}}\\ & {\mathbf{=}}& {\mathbf{Fdcos\varphi}}\end{array}$ **

Calculate the magnitude of the displacement as below.

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

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

$\u2206\mathrm{K}=+30.0\mathrm{J}$

Therefore, the angle is,

$\begin{array}{rcl}\mathrm{\varphi}& =& {\mathrm{cos}}^{-1}\left(\frac{\u2206\mathrm{K}}{\mathrm{Fd}}\right)\\ & =& {\mathrm{cos}}^{-1}\left(\frac{30.0\mathrm{J}}{\left(12.0\mathrm{N}\right)\left(5.39\mathrm{m}\right)}\right)\\ & =& {\mathrm{cos}}^{-1}\left(0.464\right)\\ & =& 62.{3}^{\xb0}\end{array}$

Hence, the angle between the force and the displacement is $62.{3}^{\xb0}$.

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

$\u2206\mathrm{K}=-30.0\mathrm{J}$

Therefore, the angle is,

$\begin{array}{rcl}\mathrm{\varphi}& =& {\mathrm{cos}}^{-1}\left(\frac{\u2206\mathrm{K}}{\mathrm{Fd}}\right)\\ & =& {\mathrm{cos}}^{-1}\left(\frac{-30.0\mathrm{J}}{\left(12.0\mathrm{N}\right)\left(5.39\mathrm{m}\right)}\right)\\ & =& {\mathrm{cos}}^{-1}\left(-0.464\right)\\ & =& 117.{64}^{\xb0}\\ \mathrm{\varphi}& \approx & {118}^{\xb0}\end{array}$

Hence, the angle between the force and the displacement is ${118}^{\xb0}$.

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