The engine of a car moving with a speed of 36 km per hour is switched off as it reaches an inclined load in client at 30 degree with the horizontal if the coefficient of friction between the road surface and the views of the car is 1 upon root 3 find the distance travelled by the car on this incline road before it comes to rest
Answers
F = normal force by the incline (due to gravity)
f = kinetic frictional force between tyres of car and surface
perpendicular to incline, force equation is given as
F = mg Cos30 ----------------------------------------E1
μ = coefficient of friction = 1/SQRT(3)
kinetic frictional force is given as
f = μF
Substituting F from E1, we get
f = μmg Cos30 -------------------------------------------E2
parallel to incline force equation is given as
mg Sin30 + f = ma
Substituting E2 into above equation, we get
mg Sin30 + μmg Cos30 = ma
a = g (Sin30 + μCos30)
a = (9.8) (Sin30 + (Cos30/√3)
a = (9.8) (1/2 + (√3/(√3*2))
a = 9.8 * (0.5 + 0.5) = 9.8
a = 9.8 m/s²
Since this is upwards, consider it as deceleration or negative acceleration.
u = initial velocity of the car = 36 km/h = 10 m/s
v = final velocity of car when it comes to stop = 0 m/s
a = acceleration = -9.8 m/s²
s = distance travelled before stopping
using the equation
v² - u² = 2as
0² = 10² + 2 (- 9.8) s
s = 5.10 m
The car comes to stop in 5.10 meters.
I also need the answer because my physics teacher may probably ask the question tomorrow