Question

problem reference 6.3
a 16.0 kg sled is being pulled along the horizontal snow - covered ground by a horizontal, constant force of 24.0 n. starting from rest, the sled attains a speed of 2.00 m/s in 8.00 m.
what is the coefficient of kinetic friction between the sled and the snow?
0.128
0.356
0.267
0.890

Explanation:

Step1: Find the acceleration

Use the kinematic equation \(v^{2}=v_{0}^{2}+2ax\).
Given \(v_{0} = 0\ m/s\), \(v = 2.00\ m/s\), \(x=8.00\ m\).
Substitute into the equation: \(2^{2}=0 + 2a\times8\).
Simplify: \(4 = 16a\), so \(a=\frac{4}{16}=0.25\ m/s^{2}\).

Step2: Apply Newton's second law

The net force \(F_{net}=ma\).
The applied force \(F = 24.0\ N\), and the frictional force \(f=\mu_{k}N\). Since \(N = mg\) (vertical equilibrium, \(N - mg=0\)), \(f=\mu_{k}mg\).
By Newton's second law \(F - f=ma\), so \(F-\mu_{k}mg=ma\).
Rearrange for \(\mu_{k}\): \(\mu_{k}=\frac{F - ma}{mg}\).
Substitute \(m = 16.0\ kg\), \(F = 24.0\ N\), \(a = 0.25\ m/s^{2}\), \(g = 9.8\ m/s^{2}\).
\(\mu_{k}=\frac{24-(16\times0.25)}{16\times9.8}=\frac{24 - 4}{156.8}=\frac{20}{156.8}\approx0.128\).

Answer:

0.128