Question

A sled slides down a snow-covered hill at constant speed. If the hillside is 10° above the horizontal, what is the coefficient of kinetic friction between the runners of the sled and the snow?​

Answer 1

Answer:

The kinetic coefficient of friction between the runners of the sled and the snow is approximately 0.176.

Explanation:

At first we present a free body diagram of the sled as an image attached below. From 1st and 2nd Newton's Laws we know that an object is at equilibrium when it is either at rest or moving at constant velocity. The equations of equilibrium associated with the sled is:

$\Sigma F_{x'} = f-W\cdot \sin \theta = 0$ (1)

$\Sigma F_{y'} = N-W\cdot \cos \theta = 0$ (2)

Where:

$\theta$ - Inclination of the hillside, measured in sexagesimal degrees.

$N$ - Normal force from ground to sled, measured in newtons.

$W$ - Weight of the sled, measured in newtons.

$f$ - Kinetic friction between sled and ground, measured in newtons.

The definitions of kinetic friction and weight are, respectively:

$f=\mu_{k}\cdot N$ (3)

$W = m\cdot g$ (4)

Where:

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$m$ - Mass of the sled, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

By applying (3) and (4) in (1) and (2), we have the following system of equations:

$\mu_{k}\cdot N-m\cdot g\cdot \sin \theta = 0$ (1b)

$N -m\cdot g \cdot \cos \theta = 0$ (2b)

And by applying (1b) in (2b), we have the following expression for the kinetic coefficient of friction:

$\mu_{k}\cdot m\cdot g \cdot \cos \theta -m\cdot g \cdot \sin \theta = 0$

$\mu_{k} = \tan \theta$ (5)

If we know that $\theta = 10^{\circ}$, then the kinetic coefficient of friction between the runners of the sled and the snow is:

$\mu_{k} = \tan 10^{\circ}$

$\mu_{k}\approx 0.176$

The kinetic coefficient of friction between the runners of the sled and the snow is approximately 0.176.