91Ó°ÊÓ

Initial Speed of car

$\left(v_i\right)=40\frac{km}{hr}$

Final speed of car

$$\begin{gathered} \left(v_f\right)=30\frac{km}{hr} \\ =8.3\frac{m}{s} \end{gathered}$$

Mass of car is 1500 kg

Distance traveled by the car 50.0 m

The incline angle is$5.0°$

Formula:

$$\begin{gathered} ∆E=∆K+∆U \\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}\left(v_f^2-v_i^2\right)\mathbf{+}\mathit{m}\mathit{g}\mathbf{∆}\mathit{y} \end{gathered}$$

Mechanical energy reduction due to friction=$-∆E$

The change in height between the car's highest and lowest points is

$$\begin{gathered} ∆y=-(50m)\sin \mathrm{θ} \\ =-4.4m \end{gathered}$$.

The lowest point (the car's final reported location) to correspond to the y = 0 reference level.

The change in potential energy is given by $∆U=mg∆y$.

As for the kinetic energy, we first convert the speeds to SI units.

$v_f=8.3\frac{m}{s}$

And

$v_i=8.3\frac{m}{s}$

The change in kinetic energy is

$∆\mathrm{K}=\frac{1}{2}m(v_f^2-v_i^2)·$

The total change in mechanical energy is $∆E=∆K+∆U$

So mechanical energy is

role="math" localid="1661405504282" $$\begin{gathered} ∆E=∆K+∆U \\ ⇒∆E=\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg∆y \\ ⇒∆E=\frac{1}{2}×1500×\left(11.1^2-8.3^2\right)+1500×9.8\left(-4.4\right) \\ ⇒∆E=-23940J \\ ⇒∆E=-2.4×{10}^{4}J \end{gathered}$$

That is, the mechanical energy reduction (due to friction) is $-2.4×{10}^{4}J$

Find $∆E_{th}$as

$∆E_{th}=f_k×d=-∆E$

Consider the distance d = 50 m.

To find magnitude of the frictional force and solve as:

$\begin{array}{l}{f}_k=\frac{∆E}{d}\\ =-\frac{\left(-2.4×{10}^4\right)}{50}\\ =4.8×{10}^{2}N\end{array}$

Magnitude of the net frictional force is $4.8×{10}^2$N.