A 1500 kgcar begins sliding down a $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{°}$inclined road with a speed of 30 km/h. The engine is turned off, and the only forces acting on the car are a net frictional force from the road and the gravitational force. After the car has traveled 50 malong the road, its speed is 40 km/h. (a) How much is the mechanical energy of the car reduced because of the net frictional force? (b) What is the magnitude of that net frictional force?

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

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

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

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

Find $∆E_{th}$as

$∆E_{th}=f_k\times d=-∆E$

Consider the distance d = 50 m.

To find magnitude of the frictional force and solve as:

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

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