A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 30 N. Starting from rest, the sled attains a speed of 2.3 m/s in 8.1 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.

A free-body diagram visualizes the forces that are acted on an object. Starting with a dot that represents the object, arrows point outward from the dot to the direction in which the forces are being acted.

In most cases, the object's gravitational or weight force points downward, while its normal force (from the surface that the object is placed on) points up.

The weight force is the product of the object's mass (in kg) and the earth's acceleration 9.81 meters per squared second.

The normal and weight forces are always equal to each other in magnitude. Because of this, the sum of forces in the y-direction for most scenarios is always 0.

Frictional forces

point in opposite directions that an applied force (propels the object) makes the object move towards;
always perpendicular to the normal force;
equals the product of the normal force and the coefficient of friction, [tex]F_f=\mu F_N[/tex].

Acceleration (Force)

There are many ways to find the acceleration of an object (like through the use of kinematics), but for this problem's sake we'll focus on the force version:

[tex]a=\dfrac{\Sigma F}{m}[/tex],

where m is the mass of the object (in kg) and the numerator portion is the sum of all forces acting upon the object (in newtons).

[tex]\dotfill[/tex]

Kinematic Equations

The basic measurements of motion are

initial/final velocity: [tex]v_0,\:v_x[/tex]
displacement: [tex]\Delta x[/tex]
time: [tex]t[/tex]
acceleration: [tex]a[/tex].

Knowing just three measurements, another measurement can be found using one (or more) of the kinematic equations

[tex]v_x=v_0+at[/tex]
[tex]\Delta x=v_0t+\dfrac{1}{2}at^2[/tex]
[tex]v_f^2=v_0^2+2a\Delta x[/tex]

(these roughly align with the AP Physics 1/2 Formula Sheet).

[tex]\hrulefill[/tex]

Solving the Problem

Drawing the FBD Diagram

We start by drawing a dot to represent the sled, and up and down arrows stemming from the dot that shows the sled's weight force and the normal force that the ground exerts.

Both will have a magnitude of 17 x 9.81 or 166.77N.

An applied force of 30N is exerted on the sled to move, we can assume the force is pointing to the right, which is usually considered as the positive direction.

This leaves the leftward direction to be the frictional force.

We can write the value of the friction force as the product of the sled's weight and mu (the coefficient of kinetic friction) [tex]F_f=166.77\mu[/tex].

(See the image attached).

From the diagram, we can figure out the sled's acceleration,

[tex]a=\dfrac{30-(166.77\mu)}{17}[/tex],

this can be verified since the sled doesn't move in the y-direction (it doesn't float or sink) but moves forward, as the problem implies.

Find Acceleration's Value Through Kinematics

The problem has given us the sled's

initial velocity [tex]v_o=0[/tex]
final velocity [tex]v_x=2.3[/tex]
displacement [tex]\Delta x = 8.1[/tex].

Looking at the third kinematics equation mentioned above, we can rearrange it to find the value of the sled's acceleration, and then plug it into the acceleration equation we made just now!

[tex]v_f^2=v_0^2+2a\Delta x[/tex]

[tex](2.3)^2=(0)^2+2a(8.1)[/tex]

[tex]2.3^2=16.2a[/tex]

[tex]0.3265=a[/tex]

Solving for the Coefficient

Plugging our acceleration value, we can rearrange the equation we've made to find the value of the coefficient.

[tex]0.3265=\dfrac{30-(166.77\mu)}{17}[/tex]

[tex]5.5505=30-(166.77\mu)[/tex]

[tex]166.77\mu=30-5.5505=24.4495[/tex]

[tex]\mu_k=0.147[/tex]

So, the coefficient of kinetic friction between the runners of the sled and the snow is 0.147.