Answer:
a) 17.53 m/s
b) 0.788
Explanation:
Let g = 9.8 m/s2. This gravitational acceleration can be split into 2 components, 1 parallel and the other perpendicular to the incline
The parallel component: [tex]g_a = gsin\theta = 9.8sin6^o = 1.024 m/s^2[/tex]
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]
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
[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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