(A) Given:Mass of block $M = 18.5 \ kg$Force $F = 40 \ N$Length of rope $L = 3 \ m$Linear density of rope $\mu = 0.5 \ kg \ m^{-1}$Distance $s = 9 \ m

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(A) Given:Mass of block $M = 18.5 \ kg$Force $F = 40 \ N$Length of rope $L = 3 \ m$Linear density of rope $\mu = 0.5 \ kg \ m^{-1}$Distance $s = 9 \ m$Mass of the rope $m_r = \mu \times L = 0.5 \times 3 = 1.5 \ kg$Total mass of the system $M_{total} = M + m_r = 18.5 + 1.5 = 20 \ kg$Acceleration of the system $a = \frac{F}{M_{total}} = \frac{40}{20} = 2 \ m \ s^{-2}$Using the equation of motion $s = ut + \frac{1}{2}at^2$,where initial velocity $u = 0$:$9 = 0 + \frac{1}{2} \times 2 \times t^2$$9 = t^2$$t = 3 \ s$

Class 11 Physics Newton's Laws of Motion and Friction Motion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact)

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$A$ block of mass $18.5 \ kg$ kept on a smooth horizontal surface is pulled by a rope of $3 \ m$ length by a horizontal force of $40 \ N$ applied to the other end of the rope. If the linear density of the rope is $0.5 \ kg \ m^{-1}$ and initially the block is at rest,the time in which the block moves a distance of $9 \ m$ is: (in $s$)

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$5$
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$7$
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Newton's Laws of Motion and Friction

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Motion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact)

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MediumAIIMS 2009

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$A$ mass $1\, kg$ is suspended by a thread. It is
$(i)$ lifted up with an acceleration $4.9\, m/s^2$
$(ii)$ lowered with an acceleration $4.9\, m/s^2$.
The ratio of the tensions is