Question

Two blocks of masses 10 kg and 2 kg are connected by an ideal spring of spring constant 1000 N/m and the system is placed on a horizontal surface as shown.

The coefficient of friction between 10 kg block and surface is 0.5 but friction is assumed to be absent between 2 kg and surface. Initially blocks are at rest and spring is unstreached then 2 kg block is displaced by 1 cm to elongate the spring then released. Then the graph representing magnitude of frictional force on 10 kg and time t. (time t is measured from that instant when 2kg is released to move)

• A.
• B.
• C.
• D. None of these

Answer 1

Answer: (A)

Answer 2

f_l = mmg = 0.5 × 10 × 10 = 50 N

max. spring force = Kx_max

= 1000 × (1 × 10^–2)N

= 10 N

so force on 10 kg block not exceed f_l so remain stationary

So a₂ = – w²x

= – w²A cos wt

So |f| = |ma₂| = |nw² A coswt|