A package is kept on a conveyor belt and the system is at rest. The belt starts to move to the right for $1.3 s$ with a constant acceleration of $2 {m/s}^{2}$ . The belt then moves with constant deceleration $a {m/s}^{2}$ and comes to a stop after a total displacement of $2.2 m$ . 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 $g = 10 {m/s}^{2}$ .

0.33 m

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0.66 m

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0.22 m

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0.11 m

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Solution

The correct option is A $0.33 m$
The FBD of the package when the belt is moving with constant acceleration $a_{1} = 2 m / s^{2}$ .

Let mass of the package be $m$
The package will be at rest relative to the belt as external force on it is $m a_{1} = 2 m$ and the limiting frictional force is $f_{l} = μ_{s} N = 0.35 m g = 3.5 m$

Now, the velocity of the package/belt is
$v = a_{1} t_{1} = 2 \times 1.3 = 2.6 m/s$
Displacement of the package/belt, when it moves with constant acceleration is
$s_{1} = \frac{1}{2} a_{1} t_{1}^{2} = \frac{1}{2} \times 2 \times (1.3)^{2} = 1.69 m$

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

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
$f_{k} = μ_{k} N = 0.25 m g = 2.5 m$

Let the package acceleration relative to the belt be $a_{r}$ .
Hence, writing the force equation

$m a_{2} - f_{k} = m a_{r}$
$\Rightarrow 6.63 m - 2.5 m = m a_{r}$
$\Rightarrow a_{r} = (6.63 - 2.5) = 4.13 {m/s}^{2}$

Thus, relative displacement is given by
$S_{r} = u_{r} t_{2} + \frac{1}{2} a_{r} t_{2}^{2} = 0 + \frac{1}{2} \times 4.13 \times (0.4)^{2}$
$= 0.33 m$

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