Question

It’s a snowy day and you’re pulling a friend along a level road

on a sled. You’ve both been taking physics, so she asks what you
think the coefficient of friction between the sled and the snow is.
You’ve been walking at a steady 1.5 m/s, and the rope pulls up on
the sled at a 30° angle. You estimate that the mass of the sled, with
your friend on it, is 60 kg and that you’re pulling with a force of
75 N. What answer will you give?

Answer 1

0.118

Step-by-step explanation:

The sled is moving at constant speed along the horizontal direction: this means that its horizontal acceleration is zero, so according to Newton's second law,

`\sum F = ma`

The net force along the horizontal direction must also be zero. Also, the sled does not move along the vertical direction, so the net force along that direction is zero as well.

The equations of the forces along the two directions are:

`F cos \theta - \mu N =0\\F sin \theta + N = mg`

where

`\theta = 30^(\circ)`

F = 75 N is the pull applied

N is the reaction force of the road on the sled

`\mu N` is the frictional force, with
`\mu` being the coefficient of friction

m = 60 kg is the total mass

g = 9.8 m/s^2

Solving for
`\mu`, we find:

`N=mg-F sin \theta\\F cos \theta - \mu(mg-F sin \theta) = 0\\\mu = (Fcos \theta)/(mg-Fsin \theta)=((75)(cos 30^(\circ)))/((60)(9.8)-(75)(sin 30^(\circ)))=0.118`