Acrate of 45.2-kg tools rests on a horizontal floor. you exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 321 n . after that you must reduce your push to 205 n to keep it moving at a steady 22.3 cm/s . part a what is the coefficient of static friction between the crate and the floor? μsμ s = nothing request answer part b

 μ = 0.725

Explanation:

This problem refers to Newton's second law.

       F = ma

Let's write the equations on each axis

Y Axis

      N-W = 0

     N = W

    N = mg

X axis

   F-fr = ma

With the body not started moving its acceleration is zero

  F-fr = 0

  F = fr

The friction force equation is

  fr = μ N

  fr = μ m g

Let's replace and calculate

   F = μ m g

   μ = F / mg

   μ = 321 /45.2 9.8

   μ = 0.725

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