Question

When an automobile moving with speed of $30km/h$ reaches an upward inclined bed boat of angle ${30}^o$ its engine is switch off if coefficient of friction envolved is $0.1$ how much difference will the automobile move before coming to rest?

Answer 1

Perpendicular to incline, force equation is given as

F = mg cos30 .....(1)

$\mu$ = coefficient of friction = 0.1

kinetic frictional force is given as

f = $\mu$F

using eq..(1)

f = $\mu$mg Cos30 .....(2)

par allel to incline force equation is given as

mg Sin30 + f = ma

using eq-(2)

mg Sin30 + $\mu$mg Cos30 = ma

a = g (Sin30 + $\mu$Cos30)

a = (9.8) (Sin30 + (0.1)Cos30)

a = 5.75 $m/s^2$

d = distance travelled before stoping

using the equation

$V^2$ = $V_0^2$ + 2 a d

$0^2$ = ${10}^2$ + 2 (-5.75) d

d = 8.7 m