Answer:
0.35
Explanation:
We resolve the component of the weight of the car along and perpendicular to the grade. We have mgsinФ and mgcosФ where Ф = angle of grade.
Now, the frictional force f = μN = μmgcosФ where μ = coefficient of friction
So, the net force along the grade is F = mgsinФ - μmgcosФ.
The work done by this force moving a distance, d along the grade is
W = (mgsinФ - μmgcosФ)d
This work equals the change in kinetic energy of the car. So ΔK = 1/2m(v₂² - v₁²) = W = (mgsinФ - μmgcosФ)d
1/2m(v₂² - v₁²) = (mgsinФ - μmgcosФ)d
1/2(v₂² - v₁²) = (gsinФ - μgcosФ)d
(v₂² - v₁²)/2d = (gsinФ - μgcosФ)
dividing through by gcosФ, we have
(v₂² - v₁²)/2dgcosФ = (gsinФ/gcosФ) - μgcosФ/gcosФ
(v₂² - v₁²)/2dgcosФ = tanФ - μ
μ = tanФ - (v₂² - v₁²)/2dgcosФ
given that tanФ = 3% = 3/100 and 1 + tan²Ф = 1/cos²Ф, cosФ = 1/(√1 + tan²Ф) = 1/(√1 + (3/100)²) = 1/(√1 + (9/10000)) = 1/(√10000 + 9/10000) = 1/√(10009/10000) = 100/√10009 = 100/100.05 = 0.9995.
Also, given that v₁ = 90 km/h = 90 × 1000/3600 m/s = 25 m/s and v₂ = 45 km/h = 45 × 1000/3600 m/s = 12.5 m/s, d = 75 m and g = 9.8 m/s².
So, substituting the values of the variables into the equation, we have
μ = tanФ - (v₂² - v₁²)/2dgcosФ
μ = 3/100 - ((12.5 m/s)² - (25 m/s)²)/(2 × 75 m × 9.8 m/s² × 0.9995)
μ = 3/100 - ((156.25 m/s)² - (625 m/s)²)/1,469.265 m²/s²
μ = 3/100 - (-468.75 m²/s²)/1,469.265 m²/s²
μ = 3/100 + 468.75 m²/s²/1,469.265 m²/s²
μ = 0.03 + 0.32
μ = 0.35
So, theoretical friction coefficient is 0.35
