If a 3/2 kg object moving at 5/3 m/s slows to a halt after moving 4/3 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Aug 20, 2017

${\mu}_{k} = 0.11$

Explanation:

We are given the following information:

• $\mapsto m = 3 / 2 \text{ kg}$
• $\mapsto \Delta s = 4 / 3 \text{ m}$
• $\mapsto {v}_{i} = 5 / 3 \text{ m"//"s}$
• $\mapsto {v}_{f} = 0$
• $\mapsto g = 9.81 {\text{ m"//"s}}^{2}$

$\implies$This problem can be solved using either kinematics and Newton's second law, or the work-energy theorem. I will include both solutions.

**$\textcolor{\mathrm{da} r k b l u e}{\text{Method 1: Newton's Second Law and Kinematics.}}$

• We can use the following kinematic equation to solve for the acceleration that the object experiences as it comes to rest:

$\textcolor{b l u e}{{v}_{f}^{2} = {v}_{i}^{2} + 2 a \Delta s}$

• After calculating the acceleration, we can generate a statement of the net force on the object using Newton's second law:

$\textcolor{b l u e}{{\vec{F}}_{\text{net}} = m \vec{a}}$

• And we can use the acceleration and given mass to calculate the net force: the force of kinetic friction ${f}_{k}$.

Let's solve for $a$ in the kinematic:

$\implies \textcolor{b l u e}{a = \frac{{v}_{f}^{2} - {v}_{i}^{2}}{2 \Delta s}}$

Using our known values:

=>a=(-(5/3" m"//"s")^2)/(2*4/3"m")

$\implies \textcolor{b l u e}{a \approx - 1.042 {\text{ m"//"s}}^{2}}$

Note that the sign of the acceleration is negative, which indicates that the acceleration is the opposite direction of motion and therefore the object is slowing down.

• As for the net force, the perpendicular (y, vertical) forces include only the normal force and force of gravity, which are in a state of equilibrium. The parallel (x, horizontal) forces, however, include only the force of kinetic friction, which is what causes the object to slow down.

${F}_{x \text{ net}} = - {f}_{k} = m a$

The statement of the perpendicular forces in equilibrium $\left(a = 0\right)$ is:

${F}_{y \text{ net}} = n - {F}_{G} = 0$

As we know that ${F}_{G} = m g :$

$\implies n = m g$

We also know that ${f}_{k} = {\mu}_{k} n$

$\implies {f}_{k} = {\mu}_{k} m g$

Therefore:

$- {\mu}_{k} m g = m a$

• Solving for ${\mu}_{k}$:

$\implies - {\mu}_{k} = \frac{m a}{m g}$

$\implies - {\mu}_{k} = \frac{a}{g}$

• Using our known values:

$- {\mu}_{k} = \left(- 1.042 {\text{ m"//"s"^2)/(9.81"m"//"s}}^{2}\right)$

$\implies \textcolor{b l u e}{0.11}$

**$\textcolor{\mathrm{da} r k b l u e}{\text{Method 2: Work-Energy Theorem.}}$

By the work-energy theorem, the work done by nonconservative forces (e.g. the force of friction) is equal to the energy lost in a system.

$\textcolor{b l u e}{\Delta {E}_{\text{sys"=W_"nc}}}$

When energy is conserved in a system, $\Delta {E}_{\text{sys}} = 0$.

We have only kinetic energy, so this statement becomes:

${W}_{\text{friction}} = \Delta K$

$\implies {W}_{F} = \frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2}$

There is no kinetic energy finally, as the object is at rest and $v = 0$:

$\implies {W}_{F} = - \frac{1}{2} m {v}_{i}^{2}$

The work done by friction:

${W}_{F} = {f}_{k} \Delta s \cos \left(\theta\right)$

• Where ${f}_{k}$ is the force of kinetic friction, $\Delta s$ is the displacement of the object, and $\theta$ is the angle between the force and displacement vectors. In this case, friction is opposite the motion and therefore antiparallel, so $\cos \left({180}^{o}\right) = - 1$

$\implies {W}_{F} = - {f}_{k} \Delta s$

$\implies - \frac{1}{2} m {v}_{i}^{2} = - {\mu}_{k} n \Delta s$

$\implies \frac{1}{2} m {v}_{i}^{2} = {\mu}_{k} m g \Delta s$

We can now solve for ${\mu}_{k}$:

$\implies {\mu}_{k} = \frac{\frac{1}{2} m {v}_{i}^{2}}{m g \Delta s}$

$\implies \textcolor{b l u e}{{\mu}_{k} = \frac{\frac{1}{2} {v}_{i}^{2}}{g \Delta s}}$

Using our known values:

$\implies {\mu}_{k} = \left(\frac{1}{2} \left(\frac{5}{3} \text{m"/"s")^2)/((9.81"m"//"s"^2)(4/3"m}\right)\right)$

$\implies {\mu}_{k} = 0.106$

$\implies \textcolor{b l u e}{{\mu}_{k} \approx 0.11}$

Aug 20, 2017

$\mu = 0.107$

Explanation:

We can find the acceleration of the object using the kinematic formula
${v}^{2} = {u}^{2} + 2 \cdot a \cdot s$
${0}^{2} = {\left(\frac{5}{3} \frac{m}{s}\right)}^{2} + 2 \cdot a \cdot \frac{4}{3} m$
Solving that for a,
$- 2 \cdot a \cdot \frac{4}{3} m = {\left(\frac{5}{3} \frac{m}{s}\right)}^{2}$
$a = - \frac{{5}^{2} \cdot 3}{2 \cdot {3}^{2} \cdot 4} {m}^{2} / \left(m \cdot {s}^{2}\right)$
$a = - \frac{{5}^{2} \cdot \cancel{3}}{2 \cdot {3}^{\cancel{2}} \cdot 4} {m}^{\cancel{2}} / \left(\cancel{m} \cdot {s}^{2}\right) = - \frac{25}{24} \frac{m}{s} ^ 2$

The force responsible for that acceleration must have been
$F = m \cdot a = \frac{3}{2} k g \cdot \left(- \frac{25}{24} \frac{m}{s} ^ 2\right)$
$F = - \frac{75}{48} N$

(I will assume that the surface this thing is on is horizontal since there is no information to tell me that I have to make it more complicated.)
The friction formula is
${F}_{f} = \mu \cdot N = \mu \cdot m \cdot g$
$- \frac{75}{48} N = \mu \cdot \frac{3}{2} k g \cdot 9.8 \frac{m}{s} ^ 2$
Solving that for $\mu$,
$\mu = - \frac{\frac{75}{48} N}{\frac{3}{2} k g \cdot 9.8 \frac{m}{s} ^ 2} = - \frac{\frac{75}{48} N}{\left(\frac{3}{2}\right) \cdot 9.8 N}$
$\mu = 0.107$
Note, the negative sign indicates that the direction of $F f$ is in the opposite direction of the initial velocity. But the coefficient of kinetic friction is a scalar, so I have dropped the sign.

I hope this helps,
Steve
p.s. double-check my arithmetic.