A block of mass 10 kg is placed on a horizontal frictionless surface and is attached to a cord which passes over two light frictionless pulleys as shown in the figure. The hanging block tied to the other end of the cord is initially at rest 2 m above the horizontal floor. If the hanging vlock strikes the floor 2 s after the system is released, then weight of the hanging block is ... `(g= 10 ms^(-2))` ,

Options

A. 22.22 N
B. 11.11 N
C. 1.11 N
D. 2.22 N

Answer

B

Explanation

For hanging block,
`s=ut+1/2 at^(2) [ because u=0] rArr 2=1/2 xx axx 2^2`
`rArr a=1ms^(-2)`
So, velocity of hanging block = velocity of 10 kg block after 2s
`rArr v=u+at `
`rArr v=0+1 xx 2` or `v=2 ms^(-1)`
From work -energy theorem, KE of 10 kg block `=1/2 xx 10 xx 2^2= mxx10xx 2`
`rArr m=1 kg`
Weight of hanging block `=1xx 10=10 kg ~~111.1N`