The system shown in figure is initially at rest. Neglecting friction determine the force 'P' required if the velocity of the collar B is 5 m/s after 2 sec and corresponding tension in the cable.

Given :

u = o m/s

v = 5m/s

t = 2 sec

$ \text{Assumptions:}$

$\text{1) The total length of the string is $L$ meter.}$

$\text{2) $x_A $and $x_B$ are the displacements of block A and B respectively.} $

$\text{3) $K_1$, $K_2$ and $K_3$ be the constant length w.r.t block A and B.} $

$\text{4) T be the tension in the string. }$

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$\text{Applying constant string length method,}$

$ \begin {aligned} L &= K_1+x_B+K_2+x_B+K_3+x_A\\ x_A &=L-K_1-2x_B-K_2-K_3 \space\space\space....(1) \end {aligned} $

$\text{Differentiating equation (1) twice w.r.t. time,}$

$a_A=-2a_B$

$\text{(We consider magnitude only)}$

$\therefore a_A=2a_B$

$\therefore a_A=2 \times (\frac{v-u}{t} )$

$\therefore a_A=2 \times (\frac{5-0}{2} )$

$\therefore a_A= 5\space m/s^2$

F.B.D of block A

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$\text{Weight of block A}$

$W_A=m_{A} \times g =1.5 \times 9.81 $

$W_A= 14.715 N $

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$\text{Applying Newtons $2^{nd} $ law of motion,}$

$\text{Along Y- axis}$

$\Sigma F_y = m_A \times a_A \\ \therefore T-W_A= m_A \times a_A \\ \therefore T-14.715=1.5 \times 5 \\ \therefore T=22.215 \space N $

F.B.D of block B

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$\text{Applying Newtons $2^{nd} $ law of motion,}$

$\text{Along X- axis}$

$ \begin {array}{I} \Sigma F_x = m_B \times a_B \\ \therefore P-2T= m_B \times a_B \\ \therefore P-(2 \times 14.715)=3 \times 2.5 \\ \therefore P=51.93 \space N \end{array}$