Box on Two Surfaces

Please let me know if I am correct or not.

Firstly the say the distance which the mass traveled be $x m$ on the rough part $x<\dfrac{2}{3}$ .

Then we can say that mass which is on the rough part= $\dfrac{Mx}{L}$ .

Now Friction acting on this part be $f_{x}=\dfrac{Mxg\mu}{L}$ .

By W-E Theorem,

$\displaystyle{\int^{L}_{0} \dfrac{Mxg\mu}{L} .dx = \dfrac{1 mv^2}{2}}$

Solving this we get

$\mu = \dfrac{3}{5} \implies \boxed{0.6}$

The energy approach is the most straightforward (see @Md Zuhair ), but here is the diff eq way as well. Say $x$ is the distance onto the rough surface. $L$ is the box length and $m$ is the box mass.

Mass of box on rough surface:

$m_r = \frac{m}{L} x$

Friction force:

$F = \mu m_r g = \frac{\mu m g}{L} x$

Newton's 2nd Law:

$-\frac{\mu m g}{L} x = m \ddot{x} \\ \ddot{x} = -\frac{\mu g}{L} x$

This obviously corresponds to simple harmonic motion of the following form (the trick, though, is that these equations only hold until the block stops):

$x = A \, cos\Big(\sqrt{\frac{\mu g}{L}} t \Big) + B \, sin\Big(\sqrt{\frac{\mu g}{L}} t \Big) \\ \dot{x} = -A \sqrt{\frac{\mu g}{L}} \, sin\Big(\sqrt{\frac{\mu g}{L}} t \Big) + B \sqrt{\frac{\mu g}{L}} \, cos\Big(\sqrt{\frac{\mu g}{L}} t \Big)$

Apply initial conditions ( $x_0 = 0, \hspace{0.5 cm} \dot{x}_0 = v_0$ ):

$A = 0 \\ B = v_0 \sqrt{\frac{L}{\mu g}}$

Revising the expressions:

$x = v_0 \sqrt{\frac{L}{\mu g}} \, sin\Big(\sqrt{\frac{\mu g}{L}} t \Big) \\ \dot{x} = v_0 \, cos\Big(\sqrt{\frac{\mu g}{L}} t \Big)$

Eventually, the velocity is zero when the cosine function becomes zero. At this point, the sine function is 1 and the distance traveled is $L$ :

$L = v_0 \sqrt{\frac{L}{\mu g}} \\ L^2 = \frac{ v_0^2 L}{\mu g} \\ \mu = \frac{v_0^2}{Lg}$

Plugging in numbers gives $\mu = \frac{3}{5}$

I think that we can do it by centre of mass concept. Centre of mass moves L/2= 1/3m Then use V^2-u^2=2as We get a=-6 It is equal to g*coeff of friction So coeff of friction= 0.6

Considering conservation of energy. The initial kinetic energy $E_k$ when the box in on the smooth surface is totally consumed as work done against friction $W$ on the rough surface.

Now, let

• $\mu$ -- the coefficient of kinetic friction between the box and the rough surface
• $m \text{ kg}$ -- the mass of the box
• $\sigma \text{ kg/m}$ -- the density per length of the box; then $\dfrac 23 \sigma = m$
• $v = 2 \text{ m/s}$ -- the velocity of the box on the smooth surface
• $s \text{ m}$ -- the displacement of the box on the rough surface

Then we have:

$\begin{aligned} W & = E_k \\ \int_0^\frac 23 \mu \sigma g s \ ds & = \frac 12 m v^2 \\ \frac {\mu \sigma g s^2}2 \ \bigg|_0^\frac 23 & = \frac 12 m v^2 \\ \frac 29 \mu \sigma g & = \frac 12 m v^2 & \small \color{#3D99F6} \text{Since }\frac 23 \sigma \\ \frac 13 \mu g & = \frac 12 v^2 \\ \implies \mu & = \frac {3v^2}{2g} = \frac {3 \times 2^2}{2 \times 10} = \boxed{0.6} \end{aligned}$

Lets consider the Energy transfers that happen on the horizontal axis.

At the beginning the box has only kinetic energy $E_{k} = \frac{mv^2}{2}$ .

At the end the box stopped due to the force of friction, so we know that it has no energy left. That is, the box did some work. That work is $A = F_{f} d = mg \mu d$ , where $d$ is the distance that the center of the mass of the box had traveled on the rough surface. Since the problem didn't state otherwise, we assume that the center of the mass of the solid rectangular box will be at the center. Hence $d = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$ meters.

Writting the equality (Energy at the beginning equal to the work at the end), we get: $\frac{mv^2}{2} = mg \mu d$ --> $\frac{mv^2}{2mgd} = \mu$ --> $\mu = \frac{v^2}{2gd}$

Plugging in the numbers: $\mu = 0.6$

After travelling 2/3 m the rectangular box stops. Also the velocity is given, so you can find out the time taken by the box to come to rest. Now we know, rate of change in momentum = applied force , which gives us the equation, (MV-0)/t = uMg , where M= Mass of the box, V= Initial velocity, t= time taken by the box ,to come to rest u= coefficient of kinetic friction , and g= 10 m/s^2. Now by plugging in the values you can find out u.

Let $L$ be the length of the block, and $x$ the length that is currently on the rough surface.

Then the friction force is $F = -\mu \frac x L F_N = -\frac{\mu m g}L x.$ As the block slides a distance $dx$ , the work done is $dW = F\:dx = -\frac{\mu m g} L x\:dx.$ The total work done by friction is $W = \int_0^L F\:dx = -\frac{\mu m g} L \int_0^L x\:dx = -\tfrac12 \mu m g L.$ This is equal to the loss in KE, so that $\tfrac12 m v_0^2 = \tfrac12 \mu m g L;$ $\mu = \frac{v_0^2}{g L} = \frac{2^2}{10\cdot \tfrac23} = \boxed{0.6}.$

Let $x$ be the length of the surface of the block that touches the rough surface. The deceleration $a$ can be written as a function of $x$ . $a= -gkx(\tfrac{2}{3})^{-1}=-15kx$ In the problem, $x$ varies uniformly from $0$ to $\tfrac{2}{3}$ .

Recall the kinematic equation: $v_f^2=v_1^2+2ad$ . In this case, $d$ is the differential unit for length equivalent to $dx$ . Plugging in our results and the given values, we must have $0=4+2\displaystyle{\int^{\tfrac{2}{3}}_{0}-15kx \,dx}$ . This yields $k=\tfrac{3}{5}$ .

Initially the rectangular box is moving over smooth where friction is assumed to be zero. So in that case there exist no friction force on it. When the box reach rough part abruptly whole length doesn't come in contact with it only small $x$ portion of $\frac{2}{3}$ of box over it. Since frictional force is directly proportional to the normal reaction as $f = \mu mg$ and friction on the $x$ will not be produced by entire mass $M$ of it rather the mass will be some portion of total mass of box which is in contact with rough surface. Small friction produced on the box is $\Delta f = \mu dm g = \frac{\mu M}{L}x$ Now total work done by frictional force is obtained by integrating from $0$ to $L$ as $\begin{aligned} & f_w =\displaystyle\int_{0}^{L} \frac{ \mu Mg }{L}x\,dx \\& = \frac{\mu Mg}{2L}L^2 \\& = \frac{\mu MgL}{2} \end{aligned}$ This work done by friction should be equal to change is kinetic energy. $\begin{aligned} & \frac{\mu MgL}{2} = \frac{1}{2}Mv^2 \\& \mu= \frac{v^2}{gL}\\& \mu = \frac{3}{5}\phantom{cccccccc}{\color{#3D99F6}\text{put L =2/3 and v = 2} }\end{aligned}$

All these answers did not notice that the distance of the box into the rough surface did not matter: friction does not depend on contact area (length).

That being the case, we have constant acceleration, and so, we are able to proceed by using simple mechanics.

$v_0 = 2\frac{m}{s}$

$v_1 = 0$

$a = -10\mu \frac{m}{s^2}$

$\Delta s = \dfrac{2}{3}m$

This because the body is at rest after traveling a distance equal to its length. We finish it off using equations of motion

$v_1^2 = v_0^2 + 2a\Delta s$

$0 = 4 - \dfrac{20}{3}\mu$

$5\mu = 3$

$\mu = \dfrac{3}{5} = 0.6$

Done.