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-frictionless-g-10-m-s-2-

Question Number 20761 by Tinkutara last updated on 02/Sep/17

A 5 k g b l o c k B i s s u s p e n d e d f r o m a

c o r d a t t a c h e d t o a 40 k g c a r t A . F i n d

t h e a c c e l e r a t i o n s o f b o t h t h e b l o c k a n d

c a r t . (A l l s u r f a c e s a r e f r i c t i o n l e s s)

(g = 10 m / s^{2})

Commented by Tinkutara last updated on 02/Sep/17

Commented by ajfour last updated on 02/Sep/17

l e t v e r t i c a l a c c e l e r a t i o n o f

b l o c k b e a, a n d a c c e l e e a t i o n o f

c a r t b e A .

m g - T = m a

T = (M + m) A

A = a

\Rightarrow m g = (M + 2 m) a

a = A = \frac{m g}{M + 2 m} = \frac{(5 k g) (10 m / s^{2})}{50 k g}

a = A = 1 m / s^{2}

n e t a c c e l e r a t i o n o f b l o c k

i s = \sqrt{2} m / s^{2} .

Commented by Tinkutara last updated on 02/Sep/17

2 doubts :

(1) Why T = (M + m) A ?

(2) Why m g = (M + 2 m) a ?

Commented by ajfour last updated on 02/Sep/17

(1) f o r t h e h o r i z o n t a l m o t i o n b o t h

c a r t a n d b l o c k c a n b e t r e a t e d a s

o n e o f m a s s (M + m) .

(2) I a d d e d t h e f i r s t t w o e q u a t i o n s

a n d w i t h a = A w e d o g e t

m g = (M + 2 m) a .

Commented by Tinkutara last updated on 02/Sep/17

Thank you very much Sir!