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

Answer:

$\mu_{k}=0.12$

Explanation:

In order to solve this problem we must first start by drawing a diagram of the situation. (See attached picture).

So we can start solving this problem by doing a sum of the forces in the y-direction so we get:

$\sum F_{y}=0$

which yields:

$N+F_{y}-W=0$

which can be solved for N so we get:

$N=W-F_{y}$

we know that:

$F_{y}=Fsin\theta$

and that

W=mg

so we can substitute that into our original equation so we get:

$N=mg-Fsin\theta$

we can also substitute the provided data here so we get:

$N=(60kg)(9.81m/s^{2})-75Nsin(30^{o})$

which yields:

N=551.1N

Next we can do a sum of forces in the x-direction so we get:

$\sum F_{x}=0$

which gives us:

$F_{x}-f=0$

We know that the friction is given by the following formula:

$f=N\mu_{k}$

and that:

$F_{x}=Fcos \theta$

so we can substitute them into our sum of forces so we get:

$Fcos \theta - N\mu_{k}=0$

we can now solve this for $\mu_{k}$:

which yields:

$\mu_{k}=\frac{Fcos \theta}{N}$

now we can substitute the provided data so we get:

$\mu_{k}=\frac{(75N)cos 30^{o}}{551.1N}=0.12$

So the coefficient of kinetic friction between the sled and the snow is 0.12.