When an automobile moving with a speed of 36km/h reaches an upward inclined road of angle 30o, its engine is switched off. If the coefficient of friction is 0.1, how much distance will the automobile move before coming to rest? Take g=10m/s2?
Answers
Answered by
6
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
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
Similar questions