A 5 kg block B is suspended from a cord attached to a 40 kg cart A . Find the accelerations of both the block and cart . ( All surfaces are frictionles ) ( g = `10 m//s^(2))`

Explanation

Forces on cart A
(1) 40 g downward
(2) normal force `n_(1)` by the floor ,
(3) Normal force `N_(2)` by the block
(4) Tension force due to string
Let acceleration of the cart is 'a' toward left then
`T_(N_(2)) = 40 a " " ` …(i)
Now let us observe the block w.r.t. cart . as string is inextensible . Acceleration of the block w.r.t. to cart will be 'a' downcard . now forces on the block are
(i) 5g downward due to gravity
(ii) Tension force due to string
(iii) `N_(2)` normal force from the cart
(iv) Pseudo force ( as block is observed w.r.t. accelerated frame ) 5a towards right
For the block ,
` N_(2) = 5a " " ` ..(ii)
` 5g - T = 5a" " ` ... (iii)
solving equation (i),(ii),(iii)
` a = 1 m//s `
So acceleration of the cart
` vec (a)_(c)= - hat (i)m//s^(2)`
And acceleration of the block w.r.t cart
` vec(a)_(Bc) = vec(a_(B))-veca_(c)`
` - hat (j) = vec (a_(B)) + hat (i)`
or , ` vec (a_(B)) = - (hat(i) +hat(j))or, a_(B) = sqrt(2) m//s^`