Question

Find the acceleration of the 500 g block in figure (5-E18).

Answer 1

Given,

$m_1=100g=0.1kg$

$m_2=500g0.5kg$

$m_3=50g=0.05kg$

From the above free body diagrams,

$T=0.5a-0.5g=0$ ...........(i)

$T_1-0.5a-0.05g=a$ ......(ii)

$T_1+0.1a-T+0.05g=0$ ..(iii)

From equation (i)

$T=0.5g-0.5a$ ........(iv)

From equation (ii)

$T_1=0.05g+0.05a$ ...(v)

Substituting equation (iv) and (v) in equation (iii), we get,

$0.05g+0.05a+0.1a-0.5g+0.5a+0.05g=0$

$⟹0.65a=0.4g⟹a=\frac{0.4}{0.65}=\frac{40}{65}g=\frac{8g}{13}downwards$

Acceleration of $4500gm$ block is $\frac{8g}{13}downwards.$