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)
Text Solution
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f λ = μ mg = 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 λ so remain stationary
So a 2 = – ω 2 x
= – ω 2 A cos ω t
So |f| = |ma 2 | = |n ω 2 A cos ω t|
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