When an automobile moving with a speed of 36km/h reaches an upward inclined road at an angle 30°, its engine is switched off. If the coefficient of friction involved is 0.1,how
tance will the automobile move before coming to rest.(g=10m/s^2).
(Show using a neat diagram the various force)
Answers
Answer:
8.7m
Explanation:
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
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