Chapter 7, Problem 25P

(a)

To determine

Speed of the projectile.

Answer to Problem 25P

The speed is $\underset{\_}{1.4\text{m}/\text{s}}$.

Explanation of Solution

Write the equation for frictional force within an isolated system.

    $\Delta K+\Delta U=-f_{k}d$        (I)

Here $\Delta K$ is the change in kinetic energy $\Delta U$ is the change in potential energy of the system, $f_k$ is the force due to friction and $d$ is the distance.

Expand (I)

  $\left(K_f-K_i\right)+\left(U_f-U_i\right)=-f_{k}d$        (II)

Here $K_f$ is the final kinetic energy, $K_i$ is the initial kinetic energy, $U_f$ is the final potential energy and $U_i$ is the initial potential energy.

Initial velocity is zero, then

  $K_i=0$        (III)

Final extension is zero, then

  $U_f=0$        (IV)

Write the equation for final kinetic energy

    $K_f=\frac{1}{2}m{v}^2$        (V)

Here, $m$ is the mass of the object and $v$ is the final velocity.

Write the equation for initial potential energy.

    $U_i=\frac{1}{2}k{x}^2$        (VI)

Here $k$ is the spring constant and $x$ is the extension of the spring.

Substitute these results in (II) and arrange in terms of $v$.

    $\begin{array}{c}-f_{k}d=\left(\frac{1}{2}m{v}^2\right)+\left(-\frac{1}{2}k{x}^2\right)\\ \left(\frac{1}{2}m{v}^2\right)=\frac{1}{2}k{x}^2-f_{k}d\\ v=\sqrt{\frac{2\left(\frac{1}{2}k{x}^2-f_{k}d\right)}{m}}\end{array}$        (VII)

Conclusion:

Substitute $8\text{N}/\text{m}$ for $f_k$, $5\text{g}$ for $m$, $0.032\text{N}$ for $f_k$, $15\text{cm}$ for $d$, $\text{5}\text{cm}$ for $x$.

    $\begin{array}{c}v=\sqrt{\frac{2\left(\frac{1}{2}\left(8\text{N}/\text{m}\right){\left(\text{5}\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2-\left(0.032\text{N}\right)\left(15\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)}{\left(5\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)}}\\ =\sqrt{\frac{2\left(\frac{1}{2}\left(8\text{N}/\text{m}\right){\left(\text{5}\times {10}^{-2}\text{m}\right)}^2-\left(0.032\text{N}\right)\left(15\times {10}^{-2}\text{m}\right)\right)}{\left(5\times {10}^{-3}\text{kg}\right)}}\\ =1.4\text{m}/\text{s}\end{array}$

The speed is $\underset{\_}{1.4\text{m}/\text{s}}$.

(b)

To determine

The point at which the ball has maximum speed.

Answer to Problem 25P

The maximum speed is achieved at $\underset{\_}{4.6\text{cm}}$ from the start.

Explanation of Solution

Write the expression for force on a spring.

    $F=kx$        (VIII)

Here $k$ is the spring constant and $x$ is the extension of the spring.

Rewrite (VI) in terms of $x$.

    $x=\frac{F}{k}$        (IX)

.Conclusion:

Substitute $0.032\text{N}$ for $F$ and $8\text{N}/\text{m}$ in (VII)

    $\begin{array}{c}x=\frac{0.032\text{N}}{8\text{N}/\text{m}}\\ =0.4\times {10}^{-2}\text{m}\\ \text{=0}\text{.400}\text{cm}\end{array}$

Therefore the ball must have moved,

  $5\text{cm-0}\text{.400}\text{cm=4}\text{.6}\text{cm}$

The maximum speed is achieved at $\underset{\_}{4.6\text{cm}}$ from the start.

(c)

To determine

Maximum speed of the projectile.

Answer to Problem 25P

The speed is $\underset{\_}{1.7\text{9}\text{m}/\text{s}}$.

Explanation of Solution

Write the equation for final potential energy

    $U_f=\frac{1}{2}k{x}^2{}_f$        (X)

Here $x_f$ is the final extension of the spring.

Substitute (III), (V), (VI0, and (X) in (II)

    $\left(\frac{1}{2}m{v}^2\right)+\left(\frac{1}{2}k{x}^2{}_f-\frac{1}{2}k{x}^2\right)=-f_{k}d$        (XI)

Rearrange (XI) in terms of $v$

    $\begin{array}{l}\left(\frac{1}{2}m{v}^2\right)=\left(\frac{1}{2}k\left(x^2-x^2{}_f\right)\right)-f_{k}d\\ v=\sqrt{\frac{2\left(\frac{1}{2}k\left(x^2-x^2{}_f\right)\right)-f_{k}d}{m}}\end{array}$

Conclusion:

Substitute $8\text{N}/\text{m}$ for $f_k$, $5\text{g}$ for $m$, $0.032\text{N}$ for $f_k$, $15\text{cm}$ for $d$, $\text{5}\text{cm}$ for $x$ and $4.6\text{cm}$ for $x_f$.

    $\begin{array}{c}v=\sqrt{\frac{2\left(\frac{1}{2}\left(8\text{N}/\text{m}\right)\left({\left(\text{5}\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2-{\left(4.6\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\right)\right)-\left(0.032\text{N}\right)\left(15\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{\left(5\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)}}\\ =\sqrt{\frac{2\left(\frac{1}{2}\left(8\text{N}/\text{m}\right)\left({\left(\text{5}\times {10}^{-2}\text{m}\right)}^2-{\left({4.610}^{-2}\text{m}\right)}^2\right)\right)-\left(0.032\text{N}\right)\left(15\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{\left(5\times {10}^{-3}\text{kg}\right)}}\\ =1.79\text{m}/\text{s}\end{array}$

The speed is $\underset{\_}{1.7\text{9}\text{m}/\text{s}}$.