Physics, asked by Arunkumar1705, 11 months ago

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

Answered by dharshuvs06
0

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

I hope that this answer helps u :-)

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Answered by indersingh1408
0
The answer will be S=8.523m
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