91Ó°ÊÓ

• Distance to wall, d = 107m.
• Mass of car, m = 1400kg.
• Initial speed, v₀ = 35m/s.
• Coefficient of static friction, ${\mathrm{μ}}_s=0.50$.

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,

$v^2=v_0^2+2ad$

The centripetal force is given by,

$F_r=\frac{m{v}_0^2}{r}$

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

$v^2=v_0^2+2ad$

Substitute the values, and we get,

$$\begin{gathered} 0={\left(35m/s\right)}^2+2a\left(107m\right) \\ a=-5.72m/s^2 \end{gathered}$$

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

$f=m\left|a\right|$

Substitute the values, and we get,

$$\begin{gathered} f=\left(1400kg\right)\left(5.72m/s^2\right) \\ f≈8.0×{10}^{3}N \end{gathered}$$

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

The maximum possible static friction is given by,

$f_{s,m}={\mathrm{μ}}_{s}mg$

Substitute the values, and we get,

$$\begin{gathered} f_{s,m}=\left(0.50\right)\left(1400kg\right)\left(9.80m/s^2\right) \\ {f}_{s,m}≈6.9×{10}^{3}N. \end{gathered}$$

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

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

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

$v=\sqrt{v_0^2+2ad}$

Substitute the values, and we get,

$$\begin{gathered} v=\sqrt{{\left(35m/s\right)}^2-2\left(0.40\right)\left(9.8m/s^2\right)\left(107m\right)} \\ v≈20m/s \end{gathered}$$

Thus, the speed at which the car will hit the wall 20m/s.

The force required to keep the motion circular is,

$F_r=\frac{m{v}_0^2}{r}$

Substitute the values, and we get,

$$\begin{gathered} F_r=\frac{\left(1400kg\right){\left(35.0m/s\right)}^2}{107m} \\ {F}_r=1.6×{10}^{4}N \end{gathered}$$

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

From the calculations, we can say that since F_r > f_s,max that means no circular path is possible.