A box of books weighing 215 N moves at a constant velocity across the floor when the box is pushed with a force of 135 N exerted downward at an angle of 30° with the horizontal. Find μk between the box and the floor.

  μk ≈ 0.414

Explanation:

The normal force at the box/floor interface is the sum of the weight and the vertical component of the applied force:

  normal force = 215 N + (135 N)sin(30°) = 282.5 N

The tangential force is the horizontal component of the applied force:

  tangential force = (135 N)cos(30°) ≈ 116.91 N

Then the coefficient of friction is ...

  μk = (tangential force)/(normal force) = (116.91 N)/(282.5 N)

  μk ≈ 0.414

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