The system is released from rest with both the spring in unstretched positions. Mass of each block is 1 kg and force constant of each spring is 10 N/m. Extension of horizontal spring in equilibrium is

Question

The system is released from rest with both the spring in unstretched positions. Mass of each block is 1 kg and force constant of each spring is 10 N/m. Extension of horizontal spring in equilibrium is, 0.2 m, 0.4 m, 0.6 m, 0.8 m

MARKS App · Accepted Answer

In equilibrium F net = 0 T = kx ........ (1) 2 T + kx 2 = mg ........ (2) Solving these equations, we have x = 2 mg 5 k = 2 × 1 × 1 0 5 × 1 0 = 0.4 m Let v be the speed of block placed horizontally, then 1 2 mv 2 + 1 2 m v 2 2 + 1 2 kx 2 + 1 2 k x 2 2 = mg x 2 or 5 8 mv 2 = mgx 2 - 5 8 kx 2 or v = 8 5 gx 2 - 5 kx 2 8 m Substituting the values, we have v = 8 5 1 0 × 0.4 2 - 5 × 1 0 × 0.16 8 × 1 = 1.26 m/s

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