91Ó°ÊÓ

• Distance to wall,$\mathrm{d}=107\text{″¾}$.
• Mass of car,$\mathrm{m}=1400\text{ k²µ}$.
• Initial speed,${\mathrm{v}}_0=35\text{″¾/²õ}$.
• Coefficient of static friction, ${\mathrm{μ}}_{\mathrm{s}}=0.50$.

The problem is based on the kinematic equations of motion in which the motion is described at constant acceleration. Also, it deals with the centripetal force. It is a force that makes a body follow a curved path.

Formula:

The velocity in kinematic equations is given by,

${\mathrm{v}}^2={\mathrm{v}}_0^2+2\mathrm{ad}$

The centripetal force is given by,

${\mathrm{F}}_{\mathrm{r}}=\frac{{\mathrm{mv}}_0^2}{\mathrm{r}}$

Using the kinematic equation, the deceleration of the car is calculated as,

${\mathrm{v}}^2={\mathrm{v}}_0^2+2\mathrm{ad}$

Substitute the values, and we get,

$\begin{array}{c}0={\left(35\mathrm{m}/\mathrm{s}\right)}^2+2\mathrm{a}\left(107\text{″¾}\right)\\ \mathrm{a}=−5.72{\text{″¾/²õ}}^{\text{2}}\end{array}$

Thus, the force of friction required to stop by car is given by,

$\mathrm{f}=\mathrm{m}\left|\mathrm{a}\right|$

Substitute the values, and we get,

$\begin{array}{l}\mathrm{f}=\left(1400\text{ k²µ}\right)(5.72{\text{″¾/²õ}}^{\text{2}})\\ \mathrm{f}≈8.0×{10}^3\text{ N}\end{array}$

Thus, the magnitude of static friction is $8.0×{10}^3\text{ N}$.

The maximum possible static friction is given by,

${\mathrm{f}}_{\mathrm{s},\mathrm{m}}={\mathrm{μ}}_{\mathrm{s}}\mathrm{mg}$

Substitute the values, and we get,

${\mathrm{f}}_{\mathrm{s},\mathrm{m}}=\left(0.50\right)(1400\text{kg})(9.80{\text{m/s}}^{\text{2}})$

${\mathrm{f}}_{\mathrm{s},\mathrm{m}}≈6.9×{10}^3\text{N}$.

Thus, the maximum possible static friction is$6.9×{10}^3\text{ N}$.

If ${\mathrm{μ}}_{\mathrm{k}}=0.40$, then${\mathrm{f}}_{\mathrm{k}}={\mathrm{μ}}_{\mathrm{k}}\mathrm{mg}$, and the deceleration is$\mathrm{a}=−{\mathrm{μ}}_{\mathrm{k}}\mathrm{g}$.

Therefore, the speed of the car when it hits the wall is,

$\mathrm{v}=\sqrt{{\mathrm{v}}_0^2+2\mathrm{ad}}$

Substitute the values, and we get,

$\begin{array}{l}\mathrm{v}=\sqrt{{\left(35\text{″¾/²õ}\right)}^2−2\left(0.40\right)\left(9.8{\text{″¾/²õ}}^{\text{2}}\right)\left(107\text{″¾}\right)}\\ \mathrm{v}≈20\text{″¾/²õ}\end{array}$

Thus, the speed at which the car will hit the wall $20\text{″¾/²õ}$.

The force required to keep the motion circular is,

${\mathrm{F}}_{\mathrm{r}}=\frac{{\mathrm{mv}}_0^2}{\mathrm{r}}$

Substitute the values, and we get,

$\begin{array}{l}{\mathrm{F}}_{\mathrm{r}}=\frac{\left(1400\text{ k²µ}\right){\left(35.0\text{″¾/²õ}\right)}^2}{107\text{″¾}}\\ {\mathrm{F}}_{\mathrm{r}}=1.6×{10}^4\text{ N}\end{array}$

Thus, the magnitude of frictional force is$1.6×{10}^4\text{ N}$ .

From the calculations, we can say that since ${\mathrm{F}}_{\mathrm{r}}>{\mathrm{f}}_{\mathrm{s},\max }$that means no circular path is possible.