Chapter 7, Problem 68P

A 1.00-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P7.68a). The object has a speed of v_i = 3.00 m/s when it makes contact with a light spring (Fig. P7.68b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Fig. P7.68c). The object is then forced toward the left by the spring (Fig. P7.68d) and continues to move in that direction beyond the spring’s unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Fig. P7.68e). Find (a) the distance of compression d, (b) the speed v at the unstretched position when the object is moving to the left (Fig. P7.68d), and (c) the distance D where the object comes to rest.

Figure P7.68

To determine

Distance of compression.

Answer to Problem 68P

The distance of compression is $\underset{\_}{0.378\text{m}}$.

Explanation of Solution

Write the energy conservation equation between second and third picture.

  $\Delta E_{\text{mech}}=\Delta K+\Delta U$        (I)

Here $\Delta E_{\text{mech}}$ is the change in internal energy due to friction, $\Delta K$ is the change in kinetic energy and $\Delta U$ is the change in potential energy.

Write the equation for change in kinetic energy,

  $\Delta K=\frac{1}{2}m{v}_f{}^2-\frac{1}{2}m{v}_i{}^2$        (II)

Here $m$ is the mass of the block, $v_f$ is the final velocity and $v_i$ is the initial velocity

Write the equation for change in potential energy,

    $\Delta U=\frac{1}{2}k{d}^2-\frac{1}{2}k{d}_i{}^2$        (III)

Here $k$ is the spring constant, $d_i$ is the initial compression distance and $d$ is the distance of compression.

The final kinetic energy zero as the final velocity zero. The initial elastic potential energy is also zero as the spring is not extended initially.

Write the equation for change in internal energy

    $\Delta E_{\text{mech}}=-\mu mgd$        (IV)

Here $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.

Substitute (II), (III) and (IV) in (I)

    $\begin{array}{c}-\mu mgd=\left(0-\frac{1}{2}m{v}_i{}^2\right)+\left(\frac{1}{2}k{d}^2-0\right)\\ 0=\frac{1}{2}k{d}^2+\mu mgd-\frac{1}{2}m{v}_i{}^2\end{array}$        (V)

Conclusion:

Substitute $50\text{N}/\text{m}$ for $k$, $0.25$ for $\mu$, $1\text{kg}$ for $m$, $3\text{m}/\text{s}$ for $v_i$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in (V) and solve the quadratic equation by eliminating the units

  $\begin{array}{l}0=\frac{1}{2}\left(50\text{N}/\text{m}\right)d^2+\left(0.25\right)\left(1\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)d-\frac{1}{2}\left(1\text{kg}\right){\left(3\text{m}/\text{s}\right)}^2\\ 0=d^2+2.45d-4.5\end{array}$

Then,

    $\begin{array}{c}d=\frac{-2.45\pm \sqrt{{2.45}^2-4\left(25\right)\left(-4.5\right)}}{2\left(2.45\right)}\\ =0.378\text{m}\end{array}$

The distance of compression is $\underset{\_}{0.378\text{m}}$.

To determine

Speed at the un stretched position when the object is moving left.

Answer to Problem 68P

The speed is $\underset{\_}{2.3\text{m}/\text{s}}$.

Explanation of Solution

Write the energy conservation equation between picture two and four

    $\Delta E_{\text{mech}}=\Delta K$        (VI)

Substitute (II) and (IV) in (VI)

    $-\mu mgd=\frac{1}{2}m{v}_f{}^2-\frac{1}{2}m{v}_i{}^2$        (VII)

Substitute $d=2d$ in (VII)

    $-\mu mg\left(2d\right)=\frac{1}{2}m{v}_f{}^2-\frac{1}{2}m{v}_i{}^2$        (VIII)

Rewrite (VIII) for $v_f$

    $v_f=\sqrt{v_i{}^2-4\mu gd}$        (IX)

Conclusion:

Substitute $3\text{m}/\text{s}$ for $v_i$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.25$ for $\mu$ and $0.378\text{m}$ for $d$ in (IX)

    $\begin{array}{c}{v}_f=\sqrt{{\left(3\text{m}/\text{s}\right)}^2-4\left(0.25\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(0.378\text{m}\right)}\\ =2.3\text{m}/\text{s}\end{array}$

The speed is $\underset{\_}{2.3\text{m}/\text{s}}$.

To determine

The distance at which the object comes to rest.

Answer to Problem 68P

The distance is $\underset{\_}{1.08\text{m}}$.

Explanation of Solution

Consider the motion from picture two to five

Substitute $D+2d$ for $d$ in (VII). Here $D$ is the distance at which the object comes to rest.

    $-\mu mg\left(D+2d\right)=\frac{1}{2}m{v}_f{}^2-\frac{1}{2}m{v}_i{}^2$        (X)

Rearrange (X) in terms of $D$.

    $\begin{array}{c}-\mu mgD-2\mu mgd=\frac{1}{2}m{v}_f{}^2-\frac{1}{2}m{v}_i{}^2\\ D=\frac{\frac{1}{2}{v}_i{}^2-\frac{1}{2}{v}_f{}^2-2\mu gd}{\mu g}\end{array}$        (XI)

Conclusion:

Substitute $3\text{m}/\text{s}$ for $v_i$, $2.3\text{m}/\text{s}$ for $v_f$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.25$ for $\mu$ and $0.378\text{m}$ for $d$ in (XI)

    $\begin{array}{c}D=\frac{\frac{1}{2}{\left(3\text{m}/\text{s}\right)}^2-\frac{1}{2}{\left(2.3\text{m}/\text{s}\right)}^2-2\left(0.25\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(0.378\text{m}\right)}{\left(0.25\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ =1.08\text{m}\end{array}$

The distance is $\underset{\_}{1.08\text{m}}$.