When an automobile moving with a speed of 36 km/hr reaches an upward inclined road of angle 30 degree, its engine is switched off. If the coefficient of friction is 0.1, how much distance will the automobile move before coming to rest?
Answers
F = normal force by the incline
f = kinetic frictional force
perpendicular to incline, force equation is given as
F = mg Cos30 eq-1
μ = coefficient of friction = 0.1
kinetic frictional force is given as
f = μF
using eq-1
f = μmg Cos30 eq-2
parallel to incline force equation is given as
mg Sin30 + f = ma
using eq-2
mg Sin30 + μmg Cos30 = ma
a = g (Sin30 + μCos30)
a = (9.8) (Sin30 + (0.1)Cos30)
a = 5.75 m/s²
consider the upward parallel to incline direction as positive
v₀ = initial velocity = 36 km/h = 10 m/s
v = final velocity when it comes to stop = 0 m/s
a = acceleration = - 5.75 m/s²
d = distance travelled before stopping
using the equation
v² = v²₀ + 2 a d
0² = 10² + 2 (- 5.75) d
d = 8.7 m
Answer:
8.5m
Explanation:
same working as below but the only difference is the exact answer is 8.5 not ----> (8.7 cannot be considered).