Question

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?

Answer 1

The coefficient of static friction is 1.13x10⁻⁵.

Step-by-step explanation:

We can find the coefficient of static friction as follows:

`-F_(\mu) = ma`

`-\mu N = ma`

`-\mu mg = ma`

`\mu = -(a)/(g)` (1)

Where:

`F_(\mu)`: 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:

`v_(f) = v_(0) + at` (2)

Where:

`v_(f)` is the final speed of the box = 0

`v_(0)` is the initial speed of the box

t is the time = 600 s

`v_(f)^(2) = v_(0)^(2) + 2aX` (3)

Where:

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

By solving equation (2) for
`v_(0)` and by entering into equation (3) we have:

`0 = (-at)^(2) + 2aX`

`a = (-2X)/(t^(2)) = (-2*20 m)/((600 s)^(2)) = -1.11 \cdot 10^(-4) m/s^(2)`

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

`\mu = -(a)/(g) = -(-1.11 \cdot 10^(-4) m/s^(2))/(9.81 m/s^(2)) = 1.13 \cdot 10^(-5)`

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

I hope it helps you!