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(\rm m)/(\rm s)\right)^2 = 2a(9.2\,\mathrm m) \implies a = 0.2875(\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 (\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 = (26.1125\,\rm N)/(166.6\,\rm N) \approx \boxed{0.16}`

(and the coefficient is dimensionless).