Answer:
3.9 m/s
Explanation:
We are given that
Mass of car,m=[tex]2.1\times 10^3 kg[/tex]
Initial velocity,u=0
Distance,s=5.9 m
[tex]\theta=19^{\circ}[/tex]
Average friction force,f=[tex]4.0\times 10^3 N[/tex]
We have to find the speed of the car at the bottom of the driveway.
Net force,[tex]F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3[/tex]
Where [tex]g=9.8 m/s^2[/tex]
Acceleration,[tex]a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}[/tex]
[tex]v=\sqrt{2as}[/tex]
[tex]v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}[/tex]
v=3.9 m/s