Chapter 6, Problem 51SP

In Fig. 6-8, h1 = 200 cm, h2 = 150 cm, and at A the 3.00-g bead has a downward speed along the wire of 800 cm/s. (a) How fast is the bead moving as it passes point-B if friction is negligible? (b) How much energy did the bead lose to friction work if it rises to a height of 20.0 cm above C after it leaves the wire?

To determine

The speed of the bead at point B if it moves at a downward velocity of $800\text{ cm/s}$ from point A in the provided diagram and the height $h_1$ is $200\text{cm}$.

Answer to Problem 51SP

Solution:

$10.2\text{ m/s}$

Explanation of Solution

Given data:

The speed at point A is $800\text{ cm/s}$.

The height $h_1$ is $200\text{ cm}$.

The height $h_2$ is $150\text{ cm}$.

Formula used:

The expression for change in kinetic energy is written as,

$KE=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $m$ is the mass, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $KE$ is the change in kinetic energy.

The expression for change in potential energy is written as,

$P{E}_G=mg\left(h_f-h_i\right)$

Here, $g$ is the acceleration due to gravity, $h_f$ is the final height, $h_i$ is the initial height, and $P{E}_G$ is the change in potential energy.

Explanation:

Consider the diagram.

Write the expression for the law of the conservation of energy.

$KE{i}_{}+P{E}_i=K{E}_f+P{E}_f$

Substitute $\frac{1}{2}m{v}_i^2$ for $K{E}_i$, $mg{h}_i$ for $P{E}_i$, $\frac{1}{2}m{v}_f^2$ for $K{E}_f$, and $mg{h}_f$ for $P{E}_f$

$\begin{array}{c}\frac{1}{2}m{v}_i^2+mg{h}_i=\frac{1}{2}m{v}_f^2+mg{h}_f\\ \frac{1}{2}{v}_i^2+g{h}_i=\frac{1}{2}{v}_f^2+g{h}_f\end{array}$

Here, $v_i$ is the speed at $A$, $h_i$ is the height at $A$, $v_f$ is the speed at $B$, $h_f$ is the height at $B$ , $m$ is the mass of the bead, and $g$ is acceleration due to gravity.

Substitute $200\text{ cm}$ for $h_1$, $0\text{ m}$ for $h_f$, $800\text{ cm/s}$ for $v_i$, and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}\frac{1}{2}{v}_f^2+\left(9.81{\text{ m/s}}^2\right)\left(0\text{ m}\right)=\frac{1}{2}{\left(800\text{ cm/s}\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)\right)}^2+\left(9.81{\text{ m/s}}^2\right)\left(200\text{ cm}\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)\right)\\ v_f^2=2\left(32+19.62\right)\\ v_f=\sqrt{103.24}\\ v_f\simeq 10.2\text{ m/s}\end{array}$

Conclusion:

The speed of the bead at point B is $10.2\text{ m/s}$.

To determine

The energy lost due to friction workif the bead rises $20\text{ cm}$ above point C in the provided diagram.

Answer to Problem 51SP

Solution:

$105\text{ mJ}$

Explanation of Solution

Given data:

The speed at point A is $800\text{ cm/s}$.

The height $h_1$ is $200\text{ cm}$.

The height $h_2$ is $150\text{ cm}$.

The mass of the bead is $3\text{ g}$.

Formula used:

The expression for change in kinetic energy is written as,

$KE=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $m$ is the mass, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $KE$ is the change in kinetic energy.

The expression for change in potential energy is written as,

$P{E}_G=mg\left(h_f-h_i\right)$

Here, $g$ is the acceleration due to gravity, $h_f$ is the final height, $h_i$ is the initial height, and $P{E}_G$ is the change in potential energy.

The expression for the law of the conservation of energy is written as,

$\begin{array}{c}{E}_i=E_f\\ KE{i}_{}+P{E}_i=K{E}_f+P{E}_f\end{array}$

Here, $E_i$ is the initial energy of the system and $E_f$ is the final energy of the system, $KE{i}_{}$ is the initial kinetic energy of the of the system, $K{E}_f$ is the final kinetic energy of the of the system, $P{E}_i$ is the initial potential energy of the of the system, and $P{E}_f$ is the final potential energy of the of the system.

Explanation:

Considered the provided diagram.

The total height covered by the bead is

$h_f=20\text{ cm}+h_2$

Here, $h_2$ is the height of point C from the horizontal.

Substitute $150\text{ cm}$ for $h_2$.

$\begin{array}{c}{h}_f=20\text{ cm}+150\text{ cm}\\ =170\text{ cm}\end{array}$

Understand that the loss of energy is converted into the work. Therefore, the mathematical expression for the same is as follows:

$K{E}_f-K{E}_i+P{E}_f-P{E}_i=W$

Substitute $\frac{1}{2}m{v}_i^2$ for $K{E}_i$, $mg{h}_i$ for $P{E}_i$, $\frac{1}{2}m{v}_f^2$ for $K{E}_i$, and $mg{h}_f$ for $P{E}_f$.

$\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg{\left(h_f-h\right)}_i=W$

Substitute $170\text{ cm}$ for $h_f$, $200\text{ cm}$ for $h_1$, $3\text{ g}$ for $m$, $800\text{ cm/s}$ for $v_i$, and $9.81{\text{ m/s}}^2$ for $g$.

$\begin{array}{c}W=\left(\begin{array}{l}\frac{1}{2}\left(3\text{ g}\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)\right)\\ \left({\left(0\text{ cm/s}\right)}^2-{\left(800\text{ cm/s}\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)\right)}^2\right)\end{array}\right)+\left(\begin{array}{l}\left(3\text{ g}\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)\right)\left(9.81{\text{ m/s}}^2\right)\\ \left(\left(170\text{ cm}\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)\right)-\left(200\text{ cm}\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)\right)\right)\end{array}\right)\\ =-0.096\text{ J}-0.008829\text{ J}\\ =-0.104829\text{ J}\left(\frac{1000\text{ mJ}}{1\text{ J}}\right)\\ \approx -105\text{ mJ}\end{array}$

Here, the negative sign denotes that the energy is being lost due to frictional work. Therefore,

Conclusion:

The energy lost due to friction work is $105\text{ mJ}$.