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)0.25 C)0.41 D)0.62 E)1.2

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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 the rate of decrease in speed, which is $0.61 \, m/s^2$. According to Newton's second law, $f = ma$, so $f_k = ma$. Given that the mass ($m$) of the puck can be found from its weight ($W = mg$), where $g = 9.8 \, m/s^2$, we have $m = \frac{W}{g} = \frac{4.0 \, N}{9.8 \, m/s^2}$. Thus, $f_k = m \cdot a = \frac{4.0}{9.8} \cdot 0.61$. The normal force ($N$) is equal to the weight of the puck ($4.0 \, N$) because it's sliding on a level surface. Substituting the values into the formula for $\mu_k$, we get $\mu_k = \frac{f_k}{N} = \frac{\frac{4.0}{9.8} \cdot 0.61}{4.0}$, which simplifies to approximately 0.062.

The Speed of a 4