The speed of a 4.0-N hockey puck, sliding across a level ice surface, decreases at the rate of 0.61 m/s². The coefficient of kinetic friction between the puck and ice is: A) 0.062 B) 0.41 C) 0.62 D) 1.2 E) 9.8

Question

Quizplus · Accepted Answer

The coefficient of kinetic friction ($\mu_k$) can be found using the formula $\mu_k = \frac{f_k}{N}$, where $f_k$ is the kinetic friction force and $N$ is the normal force. The deceleration ($a$) is given by $a = 0.61 \, \text{m/s}^2$, and using Newton's second law ($F = ma$), we can find $f_k = ma = 4.0 \, \text{N} \times 0.61 \, \text{m/s}^2 = 2.44 \, \text{N}$. The normal force ($N$) is equal to the weight of the puck, which is $4.0 \, \text{N}$ (since the weight is the force due to gravity and is given directly as $4.0 \, \text{N}$, assuming $g = 9.8 \, \text{m/s}^2$ and the mass is such that when multiplied by $g$ results in a $4.0 \, \text{N}$ force). Therefore, $\mu_k = \frac{2.44}{4.0} = 0.61$. However, there seems to be a calculation mistake in my explanation; the correct approach to find $\mu_k$ involves using the deceleration and the gravitational acceleration. The correct formula that relates these quantities is $\mu_k = \frac{f_k}{N} = \frac{ma}{mg} = \frac{a}{g}$, thus $\mu_k = \frac{0.61}{9.8} \approx 0.062$, which matches option A.

The Speed of a 4