A 15g block is sliding along horizontal plane and is given an initial velocity. The coefficient of kinetic friction between the plane and the block is 0.3. How far does the block travel before coming to a complete stop?

Explanation:

The question is incomplete without a value for the initial velocity, which i assume you meant to say it is 4.65 m/s

we will begin by applying the newtons second law of motion stated below

Given data  

v = 0

u= 4.65 m/s

m= 15 g= 0.015 kg

μ= 0.3

we know that F= ma

  a = F/m

also F= -μR

and R= mg

i.e F= μmg

hence

a= -μmg/m = -μg  [minus sign because the acceleration is opposite the direction of motion]

Making s subject of formula we have

0 - u^2 = 2(-μg)s

s = U^2 / (2μg)

Substituting in the second equation of motion we have

 The block will go 3.67 meters before stopping

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s