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A package is at rest on a conveyor belt which is initially at rest. The belt is started and moves to the right for 1.3 seconds with a constant acceleration of $2 m / s^{2}$ . The belt then moves with a constant deceleration $a_{2}$ and comes to a stop after a total displacement of 2.2m. Knowing that the coefficients of friction between the package and the belt are $μ_{} = 0.35$ and $μ_{k} = 0.25$ , determine the deceleration $a_{2}$ of the belt,

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$5.63 m / s^{2}$

$4.63 m / s^{2}$

$6.63 m / s^{2}$

$7.63 m / s^{2}$

answer is C.

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Detailed Solution

$v = a_{1} t_{1} = 2.6 m / s$

$s_{1} = \frac{1}{2} a_{1} t_{1}^{2} = \frac{1}{2} \times 2 \times (1.3)^{2} = 1.69 m s_{2} = (2.2 - 1.69) = 0.51 m$

Now, $s_{2} = \frac{v^{2}}{2 a_{2}}$

$∴ a_{2} = \frac{v^{2}}{2 s_{1}} = \frac{(2.6)^{2}}{2 \times 0.51} = 6.63 m / s^{2}$

and

$t_{2} = \frac{v}{a_{2}} = 0.4 s$

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