Question

A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 31 N. Starting from rest, the sled attains a speed of 2.3 m/s in 9.2 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
What is the number and units?

Answer 1

The sled accelerates with magnitude $a$ such that

$\left(2.3\dfrac{\rm m}{\rm s}\right)^2 = 2a(9.2\,\mathrm m) \implies a = 0.2875\dfrac{\rm m}{\mathrm s^2}$

By Newton's second law, the net force in the plane of motion (parallel to the ground) is

$31\,\mathrm N - F_{\rm friction} = (17 \,\mathrm{kg}) \left(0.2875 \dfrac{\rm m}{\mathrm s^2}\right)$

so that the force of friction exerts a magnitude of

$F_{\rm friction} = 26.1125 \,\mathrm N$

Perpendicular to the ground, the sled is in equilibrium, so Newton's second law says

$F_{\rm normal} - (17\,\mathrm{kg})g = 0 \implies F_{\rm normal} = 166.6 \,\mathrm N$

The magnitude of friction is proportional to the magnitude of the normal force by a factor of $\mu_k$, the coefficient of kinetic friction. It follows that

$F_{\rm friction} = \mu_k F_{\rm normal} \implies \mu_k = \dfrac{26.1125\,\rm N}{166.6\,\rm N} \approx \boxed{0.16}$

(and the coefficient is dimensionless).