A box is sliding on a rough patch of ice, and the only horizontal force acting on it is friction. You observe that it slides to a stop in 600 seconds over a distance of 20 meters. What magnitude is coefficient of static friction between the box and the ice?

The coefficient of static friction is 1.13x10⁻⁵.      

Explanation:

We can find the coefficient of static friction as follows:

   (1)

Where:

: is the friction force

m: is the mass of the box

a: is the acceleration

g: is the gravity = 9.81 m/s²

First, we need to calculate the acceleration:

     (2)

Where:

is the final speed of the box = 0

is the initial speed of the box

t is the time = 600 s

   (3)

Where:

X: is the distance traveled by the box = 20 m

By solving equation (2) for and by entering into equation (3) we have:

Now, we can calculate the coefficient of static friction by entering the above value into equation (1) :

Therefore, the coefficient of static friction is 1.13x10⁻⁵.              

I hope it helps you!                                                                                  

(75°f − 32) × 5/9 = 23.889°c

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day