Question

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?

Short answer

1. Reduced mechanical energy of the car is$2.4\times {10}^4\hspace{0.17em}J$
2. Magnitude of the net frictional force is$4.8\times {10}^2\hspace{0.17em}N$

Step-by-step solution

Given

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$5.0°$

Determine the concept and the formulas:

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=$-∆E$

(a) Calculate how much the mechanical energy of the car reduced because of the net frictional force

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 $-2.4\times {10}^{4}J$

(b) Calculate the magnitude of that net frictional force

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.