A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an average friction force of 4.0 ✕ 103 N impedes the motion, find the speed of the car at the bottom of the driveway.

Respuesta :

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

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