Chapter 5, Problem 104A

To determine

To Calculate: The distance travelled by the car before it stops.

Answer to Problem 104A

The distance travelled by the car before it stops is $49.05\text{ m}$ .

Explanation of Solution

Given info:

The initial distance from which disk started, $d_i=0\text{m}$

The initial velocity of the disk, $u=25\text{m/s}$

The kinetic coefficient of friction between disk and surface, ${\mu }_k=0.65$

The acceleration due to gravity, $g=9.81{\text{m/s}}^{\text{2}}$ .

Formula used:

From the Newton’s second law:

  $\begin{array}{l}{\displaystyle \sum F}=-ma\\ F_f=-ma\\ {\mu }_{k}mg=-ma\\ a=-{\mu }_{k}g\end{array}$

Here, $F_f$ represents force of friction.

  $m$ represents the mass of the disk.

  $a$ represents the acceleration of the disk.

  ${\mu }_k$ represents the coefficient of kinetic friction between disk and surface.

From the Newton equation of motion:

  $\begin{array}{l}{v}^2-u^2=2as\\ s=d_f-d_i=d_f\\ d_f=\frac{v^2-u^2}{2a}\end{array}$

Here, $v$ represents final velocity of the disk.

  $s$ represents the distance traveled.

  $a$ represents the acceleration due to gravity of the disk.

  $u$ represents the initial velocity of the disk.

  $d_f$ represents the final distance to which disk moved.

  $d_i$ represents the initial distance from which disk started.

  $Q=mL$

Here, L is the latent heat of fusion of water in J/gm.

Calculation:

Substituting the given values,

  $\begin{array}{c}a=-{\mu }_{k}g\\ =-65\times 9.81\\ =-9.32{\text{m/s}}^2\end{array}$

From the Newton equation of motion:

  $\begin{array}{c}{d}_f=\frac{v^2-u^2}{2a}\\ v=0\text{m/s}\\ u=25\text{m/s}\\ a=-9.32{\text{m/s}}^{\text{2}}\\ d_f=\frac{0^2-{25}^2}{2(-6.37)}\\ =\frac{625}{12.74}=49.05\text{m}\end{array}$

Conclusion:

The distance travelled by the car before it stops is $49.05\text{m}$ .