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Found in: Page 238

### Physics For Scientists & Engineers

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

# The coefficient of friction between the block of mass ${{\mathbit{m}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.00}}{\mathbf{}}{\mathbf{}}{\mathbit{k}}{\mathbit{g}}$and the surface in Figure P8.22 is.${{\mathbf{\mu }}}_{k}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{400}}$ The system starts from rest. What is the speed of the ball of mass ${{\mathbit{m}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.00}}{\mathbf{}}{\mathbf{}}{\mathbit{k}}{\mathbit{g}}$when it has fallen a distance${\mathbf{h}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{5}}{\mathbf{}}{\mathbf{}}{\mathbf{m}}$?

The speed of the ball when it has fallen a distance $h=1.5\text{m}$ is$v=3.74\text{\hspace{0.17em}m/s}$ .

See the step by step solution

## Step 1: Law of conservation of energy

In the presence of dissipative forces, the law of conservation of energy states that the total change in mechanical energy during a process, should be equal to the increase in internal energy of the system.

$\begin{array}{c}{\mathbf{\Delta E}}_{\mathrm{mech}}\mathbf{=}\mathbf{\Delta K}\mathbf{+}{\mathbf{\Delta U}}_{g}\\ \mathbf{=}\mathbf{-}{\mathbf{f}}_{k}\mathbf{d}\\ \mathbf{=}\mathbf{-}{\mathbf{\Delta E}}_{\mathrm{int}}\text{\hspace{0.17em}\hspace{0.17em}}...............\left(\mathbf{8}\mathbf{.}\mathbf{16}\right)\end{array}$

If a system is not isolated as well, the total change in energy is given as-

$\begin{array}{c}\mathbf{\Delta E}\mathbf{=}\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{+}{\mathbf{\Delta E}}_{\mathrm{int}}\\ \mathbf{=}\sum {W}_{\mathrm{other}\mathrm{forces}}\text{\hspace{0.17em}\hspace{0.17em}}\mathbf{.}\mathbf{...........}\mathbf{\left(}\mathbf{8}\mathbf{.}\mathbf{17}\mathbf{\right)}\end{array}$

here,

${\mathbf{\Delta K}}{=}$ Change in kinetic energy

${\mathbf{\Delta U}}{=}$ Change in potential energy

${{\mathbf{\Delta E}}}_{\mathrm{int}}{=}$ Change in internal energy

## Step 2: Given Data

mass of the block: ${m}_{1}=3.00\text{\hspace{0.17em}kg}$

mass of the ball: ${m}_{2}=5.00\text{\hspace{0.17em}kg}$

coefficient of friction: ${\mu }_{k}=0.400$

distance moved: $h=1.5\text{\hspace{0.17em}m}$

## Step 3:

Use Equation 8.16:

$\begin{array}{c}\Delta {E}_{mech}=\Delta K+\Delta {U}_{g}\\ =-{f}_{k}d\\ =-\Delta {E}_{\mathrm{int}}\end{array}$

For the system of block, ball and surface; let the subscript ‘1’ and ‘2’ denote the values of block and ball respectively. So, equation becomes-

data-custom-editor="chemistry" $\Delta {K}_{1}+\Delta {U}_{1}+\Delta {K}_{2}+\Delta {U}_{2}=-{f}_{k}d$

Choose the initial point before release and the final point after each block have moved$1.5\text{\hspace{0.17em}}m$ . Choose$u=0$ with the$3.00\text{\hspace{0.17em}}kg$ block on the tabletop and the block in its final position.

So ${K}_{i1}={K}_{i2}={U}_{1f}={U}_{2f}={U}_{i1}=0$

We now have

$\frac{1}{2}{m}_{1}{v}_{f}^{2}+\frac{1}{2}{m}_{2}{v}_{f}^{2}+0+0-0-0-0-{m}_{2}g{y}_{2i}=0-{f}_{k}d$

where the friction force is

$\begin{array}{c}{f}_{k}={\mu }_{k}n\\ ={\mu }_{k}{m}_{a}g\end{array}$

The friction force causes a negative change in mechanical energy because the force opposes the motion. Since all of the variables are known except for,${v}_{f}$ we can substitute and solve for the final speed.

$\frac{1}{2}{m}_{1}{v}_{f}^{2}+\frac{1}{2}{m}_{2}{v}_{f}^{2}-{m}_{2}g{y}_{2i}=-{f}_{k}d$

$\begin{array}{l}{v}^{2}=\frac{2gh\left({m}_{2}-{\mu }_{k}{m}_{1}\right)}{{m}_{1}+{m}_{2}}\\ v=\sqrt{\frac{2\left({\text{9.8 m/s}}^{\text{2}}\right)\left(\text{1.5 m}\right)\left[5.\text{00 kg}-\text{0.400}\left(\text{3.00 kg}\right)\right]}{8.00\text{\hspace{0.17em}}\text{kg}}}\\ v=3.74\text{m/s}\end{array}$

After falling $1.5\text{\hspace{0.17em}m}$, the ball is moving with a velocity of .$3.74\text{m/s}$