A trolley of mass 5,kg on a horizontal smooth | vedclass

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(B) The total mass of the system is $M_{total} = 5\,kg$ (trolley) $+ 1\,kg$ (rope) $+ 2\,kg$ (load) $= 8\,kg$.When $BC = 0$,the entire mass of the rop

Vedclass · Accepted Answer

$\frac{20}{8}$ to $\frac{30}{8}$

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(B) The total mass of the system is $M_{total} = 5\,kg$ (trolley) $+ 1\,kg$ (rope) $+ 2\,kg$ (load) $= 8\,kg$.When $BC = 0$,the entire mass of the rope is on the horizontal surface. The driving force is the weight of the load,$F = m_{load} \cdot g = 2g$.The acceleration is $a_1 = \frac{F}{M_{total}} = \frac{2g}{8} = \frac{20}{8}\,m/s^2$.When $BC = 2\,m$,the entire rope is hanging vertically. The driving force is the weight of the load plus the weight of the rope,$F = (m_{load} + m_{rope}) \cdot g = (2 + 1)g = 3g$.The acceleration is $a_2 = \frac{F}{M_{total}} = \frac{3g}{8} = \frac{30}{8}\,m/s^2$.Thus,the acceleration changes from $\frac{20}{8}\,m/s^2$ to $\frac{30}{8}\,m/s^2$.

$A$ trolley of mass $5\,kg$ on a horizontal smooth surface is pulled by a load of mass $2\,kg$ by means of a uniform rope $ABC$ of length $2\,m$ and mass $1\,kg$. As the load falls from $BC = 0$ to $BC = 2\,m$,its acceleration in $m/s^2$ changes from:

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