Question

A package is kept on a conveyor belt and the system is at rest. The belt starts to move to the right for $1.3\text{s}$ with a constant acceleration of $2{\text{m/s}}^2$. The belt then moves with constant deceleration $a{\text{m/s}}^2$ and comes to a stop after a total displacement of $2.2\text{m}$. Knowing that the coefficient of static friction between the package and the belt is $0.35$ and coefficient of kinetic friction is $0.25$. Then determine the displacement of the package relative to the belt as the belt comes to stop. Take $g=10{\text{m/s}}^2$.

• A. $0.33\text{m}$
• B. $0.66\text{m}$
• C. $0.22\text{m}$
• D. $0.11\text{m}$

Answer 1

The correct option is A $0.33\text{m}$

The FBD of the package when the belt is moving with constant acceleration $a_1=2m/s^2$.


Let mass of the package be $m$
The package will be at rest relative to the belt as external force on it is $m{a}_1=2m$ and the limiting frictional force is $f_l={\mu }_{s}N=0.35mg=3.5m$

Now, the velocity of the package/belt is
$v=a_1{t}_1=2\times 1.3=2.6\text{m/s}$
Displacement of the package/belt, when it moves with constant acceleration is
$s_1=\frac{1}{2}{a}_1{t}_1^2=\frac{1}{2}\times 2\times (1.3{)}^2=1.69\text{m}$

Given, total displacement is $2.2\text{m}$
Now, the displacement of the belt during retardation is
$s_2=(2.2-1.69)=0.51\text{m}$
Also, from the equation of motion
$s_2=\frac{v^2}{2a_2}$
$\therefore a_2=\frac{v^2}{2s_2}=\frac{(2.6{)}^2}{2\times 0.51}=6.63{\text{m/s}}^2$
and $t_2=\frac{v}{a_2}=\frac{2.6}{6.63}=0.4\text{s}$

Relative motion will start only when the belt starts to decelerate.
So, the friction force acting on the package will be kinetic in nature, which is given by
$f_k={\mu }_{k}N=0.25mg=2.5m$

Let the package acceleration relative to the belt be $a_r$.
Hence, writing the force equation


$m{a}_2-f_k=m{a}_r$
$\Rightarrow 6.63m-2.5m=m{a}_r$
$\Rightarrow a_r=(6.63-2.5)=4.13{\text{m/s}}^2$

Thus, relative displacement is given by
$S_r=u_r{t}_2+\frac{1}{2}{a}_r{t}_2^2=0+\frac{1}{2}\times 4.13\times (0.4{)}^2$
$=0.33\text{m}$