Two blocks`A` and`B` are connected to each other by a string and a spring , the spring pases and a frictionlesss pulley as down over in the figure . Block `B` sides over the horizental top surface of a strionry block `C` both with the verical side of `C` , both with the same conform speed

The coefficient of friction between the surface the of block is `0.2` force constant of the spring is `1960` newtons , if mass of block `A` is `2` kg , celculate the mass of block `F` and `B` and the energy stored is the spring

Answer

A

Explanation

Let m be the mass of B. From its free-body diagram
`T-muN=m xx 0 =0`
.
where, `T="tension" of the string and `N=mg`
`:. T=mumg`
From the free-body diagram of the spring
`T-T'=0`
where, T' is the force exerted by A on the spring or `T'=mumg`
From the free-body diagram of A `2g-(T' + muN') =2xx0=0`
where, N' is the normal reaction of the vertical wall of C on A and `N' + 2 xx 0` (as there is no horizontal acceleration of A)
Now, in a spring tensile force=force constant xx enxtension
`:. 19.6 =1960x` or `x=(1)/(100)m` or `U("energy of a spring") =(1)/(2)kx^(2)`
`=1/2 xx 1960 xx ((1)/(100))^(2) =0.098 J`