Answer:
μ = 0.352
Note: This answer was rounded to three significant figures.
Explanation:
To determine the coefficient of kinetic friction for the crate, we need to use the given information about the frictional force and the deceleration of the crate.
We are given:
- f_k = 78.0 N
- a = -3.45 m/s²
Summing the horizontal forces acting on the crate gives us the equation:
[tex]\Longrightarrow -\vec f_k = m\vec a[/tex]
The equation for the kinetic friction forces is given below, plug this into the equation above.
[tex]\boxed{ \begin{array}{ccc} \text{\underline{Kinetic Frictional Force:}} \\\\ \vec f_k = \mu_k \vec n \\\\ \text{Where:} \\ \bullet \ \vec f_k \ \text{is the kinetic frictional force} \\ \bullet \ \mu_k \ \text{is the coefficient of kinetic friction} \\ \bullet \ \vec n \ \text{is the normal force} \end{array} }[/tex]
[tex]\Longrightarrow -\mu \vec n = m\vec a[/tex]
Solve for 'μ':
[tex]\therefore \mu = -\dfrac{m\vec a}{\vec n}[/tex]
Since there are no other vertical forces applied to the crate, the normal force is equal to the weight of the crate:
[tex]\boxed{ \begin{array}{ccc} \text{\underline{Weight of an Object:}} \\\\ \vec w = mg \\\\ \text{Where:} \\ \bullet \ \vec w \ \text{is the weight of the object (force due to gravity)} \\ \bullet \ m \ \text{is the mass of the object} \\ \bullet \ g \ \text{is the acceleration due to gravity} \end{array} }[/tex]
[tex]\Longrightarrow \mu = -\dfrac{m\vec a}{\vec w}\\\\\\\\\Longrightarrow \mu = -\dfrac{m\vec a}{mg}\\\\\\\\\therefore \mu=-\dfrac{\vec a}{g}[/tex]
Plugging in our values gives us a result of:
[tex]\Longrightarrow \mu=-\dfrac{-3.45 \text{ m/s}^2}{9.8 \text{ m/s}^2} \approx \boxed{0.352}[/tex]
Thus, the coefficient of kinetic friction is approximately 0.352.
