A 2000kg car starts from rest at point A on a 6 degree incline and coasts through a distance of 150m to point B. The brakes are then applied, causing the car to come to a complete stop at point C, 20m from B. Knowing that slipping is impending during the breaking period and neglecting air resistance and rolling resistance, determine (a) The speed of the car at point B, (b) the coefficient of static friction between the tires and the road.

We can use the following equation of motion to find the velocity at point B, knowing that it starts from rest at point A and coast through a distance of s = 150 m at rate of 1.024 m/s2

[tex]v_B^2 - v_A^2 = 2g_as[/tex]

[tex]v_B^2 - 0^2 = 2*1.024*150[/tex]

[tex]v_B^2 = 307.3[/tex]

[tex]v_b = \sqrt{307.3} = 17.53 m/s[/tex]

b) we can apply the same formula to find out the deceleration from B to C, knowing that it comes to a stop at C after S = 20m

[tex]v_C^2 - v_B^2 = 2a_fS[/tex]

[tex]0^2 - 307.3 = 2*a_f*20[/tex]

[tex]a_f = \frac{-307.3}{40} = -7.683 m/s^2[/tex]

Using Newton's 2nd law we can find the friction force

[tex]F_f = a_fm = 7.683*2000 = 15365 N[/tex]

Which consists of normal force N and coefficient of static friction [tex]\mu_f[/tex]

[tex]F_f = \mu_fN = \mu_fmgcos\theta[/tex]

[tex]\mu_f = \frac{F_f}{mgcos\theta} = \frac{15365}{2000*9.8cos6^o} = 0.788[/tex]

Histrionicus Histrionicus · Answer 2 · 2021-12-15T06:16:11+01:00

In the given case, (a) The speed of the car at point B - 17.53

(b) the coefficient of static friction between the tires and the road - 0.788

We know g = 9.8 m/s2 that is gravitational acceleration can be split into 1 parallel and the other perpendicular components to the incline

The parallel component:

[tex]g_a = gsin\theta = 9.8sin6^o = 1.024 m/s^2[/tex]

This equation of motion to find the velocity at point B, knowing that it starts from rest at point A and coasts through a distance of s = 150 m at a rate of 1.024 m/s²

[tex]v_B^2 - v_A^2 = 2g_as\\v_B^2 - 0^2 = 2*1.024*150\\\\v_B^2 = 307.3\\v_b = \sqrt{307.3} = 17.53 m/s[/tex]

b) to find out the deceleration from B to C, knowing that it comes to a stop at C after S = 20m

[tex]v_C^2 - v_B^2 = 2a_fS\\0^2 - 307.3 = 2\times a_f\times 20\\a_f = \frac{-307.3}{40} = -7.683 m/s^2[/tex]

Using Newton’s 2nd law we can find the friction force

[tex]F_f = a_fm = 7.683\times 2000[/tex]

= 15365 N

Which consists of normal force N and coefficient of static friction

[tex]F_f = \mu_fN = \mu_fmgcos\theta[/tex]

[tex]\mu_f = \frac{F_f}{mgcos\theta}[/tex]

=[tex]\frac{15365}{2000*9.8cos6^o}[/tex]

= 0.788

Thus, In the given case, (a) The speed of the car at point B - 17.53

(b) the coefficient of static friction between the tires and the road - 0.788

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