I came across this this old Applied Maths problem and need confirmation that my logic is correct.
A parcel rests on a stationary conveyor belt which slopes at an angle of $30^{\circ}$ to the horizontal. The belt begins to move upwards with an increasing acceleration and when this reaches $0.7\,m/s^2$ the parcel begins to slip. Calculate the coefficient of friction between the parcel and the belt. If, before the parcel begins to slip, the belt is suddenly stopped when its speed reaches $2\frac{1}{7} \,m/s$, how far along the belt would the parcel slide?
I've successfully found the coefficient of friction between the parcel and the belt. Using Newton's Second Law, and taking moving up the plane as positive. The component of the mass down the plane is $mg \sin 30^{\circ}$. Friction acts up the plane against the motion of the conveyor belt. $N$ is the normal reaction of mass on the plane.
$$ \mu N - mg \sin 30^{\circ} = ma $$
Since $N = mg \cos 30^{\circ}$. \begin{align} \\& \mu mg \cos 30^{\circ} - mg \sin 30^{\circ} = m \frac{7}{10} \\& \mu g \frac{\sqrt{3}}{2} - \frac{g}{2} = \frac{7}{10} \\& \mu = \frac{11.2}{g \sqrt{3}} \\& \mu \approx 0.66 \end{align} This answer is correct according to the book. My problem is the second part of the question. My logic is that when the conveyor belt belt stops the particle has an initial speed of $2\frac{1}{7} \,m/s$, I can find the acceleration (or deceleration in this case), then apply $v^2 = u^2 + 2as$.
Equations of motion is still up then plane, however the friction acting on the particle is now down the plane.
$$ - mg \sin 30^{\circ} - \mu N = m a $$ $$ - mg \sin 30^{\circ} - 0.66 mg \cos 30^{\circ} = m a $$
$$ -\frac{g}{2} - 0.66 g \frac{\sqrt{3}}{2} = a $$
$$ a = -10.5 $$
Using $v^2 = u^2 + 2as$. Particle comes to rest after traveling $s$ meters.
$$ 0 = \left(\frac{15}{7} \right)^2 + 2(-10.5)s $$
$$ s = 0.219 $$
The book gives the answer as $0.44$.
Which seems the be half of the book's answer. Any help appreciated.
I have come back to this problem and discovered the solution.
Using the deceleration of $10.5\,m/s^2$.
$$ a = \dfrac{v}{t} $$ $$ 10.5 = \dfrac{\frac{15}{7}}{t} $$ $$ t = 0.2041 $$
The time taken the for the distance covered is $0.2041\,s$. Using $v = \dfrac{s}{t}$.
$$ s = v \cdot t $$ $$ s = \frac{15}{7} \cdot 0.2041 $$ $$ s \approx 0.44 $$