Two blocks A and B are connected to each other by a string and a spring, the string passes over a frictionless pulley as shown in Block B slides over the horizontal surface of a stationary block C and the block. A slides along the vertical side of C, both with same uniform speed. The coefficient of friction between the surface of the blocks is 0.2 Force constant of the spring is 1960 `N//m` if the mass of block A is 2 kg calculate the mass of the block B and the energy stored in the spring.

Explanation

In this problem:
(1) As masses A and B are moving with constant velocity this is a problem of dynamic equilibium i.e., forces acting on mass A (or B) balance each other.
(2)As string and spring are weithless and no mass is involved between them, `T_("string")=T_("spring")=T`
(3) Force of friction on block B, `f_(B)=muR_(B)=mum_(B)g` (as `R_(B)=m_(B)g)` while on block A, `f_(A)=muR_(A)=0` (as `R_(A)=m_(A) g cos 90^(@)=0)`
In the light of above for horizontal equilibrium of B,
`T=f_(B)=mum_(B)g` ....(i)
While for vertical equilibrium of A,
`T=m_(A)g` ....(ii)
So, from Eqns (i) and (ii)
`mum_(B)g=m_(A)g`
`m_(B)=(2//0.2)=10kg`

Now, as for spring `T=ky` in the light of Eqn. (ii) becomes
`ky=m_(A)g i.e., y=(2xx9.8)//1960=10^(-2)m`
So, the enegy stored in the spring
`U=(1//2)ky^(2)=(1//2)xx1960xx(10^(-2))^(2)=0.098J`