Question

A 20-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 26 n. starting from rest, the sled attains a speed of 2.3 m/s in 8.9 m. find the coefficient of kinetic friction between the runners of the sled and the snow.

Answer 1

Good morning.

We calculate the acceleration with the Torricelli equation:

$\star\ \boxed{\mathsf{V^2 = V_0^2 + 2a\Delta S}}$

We see that:

$\begin{cases}\mathsf{V_0 = 0}\\\mathsf{V = 2.3 \ m/s}\\\mathsf{\Delta S = 8.9 \ m}\end{cases}$

Now:

$\mathsf{(2.3)^2 = 0^2 + 2a\cdot 8.9}\\ \\ \mathsf{5.29 = 17.8a}\\ \\ \bold{\mathsf{a = 0.3 \ m/s^2}}$


Now we can calculate the resultant force that makes that acceleration of 0.3 m/s² with the 2nd Law of Newton:

$\mathsf{F_r = m\cdot a}\\ \\ \mathsf{F_r = 20\cdot 0.3}\\ \\ \mathsf{F_r = 6 \ N}$


We have a force of 26 N → and a friction force F ←. Adding those vectors, he have  a force 6 N →. Therefore:

26 - F = 6

F = 20 N


We have a friction force of 20 N. We calculate the kinect coefficient with the formula:

$\star \ \boxed{\mathsf{F = \mu_k N}}$


Since we are in a horizontal plane, we hava that N = P = mg = 200 N

Therefore:

$\mathsf{20 = \mu_k 200}\\ \\ \boxed{\boxed{\mathsf{\mu_k = 0.1}}}$