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Four packages each having a mass of $4 kg$ are attached on the belt at equal distances $d = 200 mm$ as shown in the figure. Initially belt is at rest. If a constant force of magnitude $840 N$ is applied to the belt, determine the velocity (in $m / s$ ) of package 2 as it falls off the belt at point A. Assume that the mass of the belt and pulleys is small compared to the mass of packages. Assume that radius of pulley is negligible in comparison to width $d$ .

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answer is 7.

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Detailed Solution

$Fd - \frac{1}{2} \times 4 \times v^{2} \times 4 \Rightarrow v^{2} = \frac{Fd}{8}$

$Fd = \frac{1}{2} \times 3 (v_{1}^{2} - v^{2}) \times 4 v_{1}^{2} = \frac{Fd}{6} + \frac{Fd}{8}$

$= \frac{7 Fd}{24} = \frac{7 \times F \times d}{24} = \frac{7 \times 840 \times 0.2}{24} = 49 \Rightarrow v_{1} = 7 m / s$

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