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 the figure. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of blocks is 0.2. Force constant of the spring is 1960 newtons/m. If mass of block A is 2 kg. The mass of block B and the energy stored in the spring are _____ kg and _____ J.

Answer

(10, 0.098)

Explanation

Blocks A and B are moving with constant velocity, therefore the net force acting on A and B should be zero. Since the spring is loaded, it will be in a deformed state. Let the extension of the spring be x.

Because A is not pushing C, hence there is no normal reaction on A. It means there will be no friction force between block A and C.
Both A and B are moving with uniform speed. Therefore, net force on them should be zero.
Fore equilibrium of A:
`m_Ag = kx :. x = (m_Ag)/(k) = ((2)(9.8))/(1960) = 0.01 m`
For equilibrium of B:
`mu m_B g = T = kx = m_Ag`
`:. m_B = (m_A)/(mu) = 2/(0.2) = 10 kg`
Energy stored in spring
`U = 1/2 kx^2 = 1/2 (1960)(0.01)^2 = 0.098 J`.