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AP EAMCET20228 Jul 2022Morning ShiftPhysicsWork, Power and EnergyActual

The work done in stretching a spring of natural length 25 cm and spring constant 50 Nm ⁻¹ from 50 ~cm to 60 ~cm is

Options

  1. A1.5 ~J
  2. B2 ~J
  3. C3.5 ~J
  4. D5 ~J

Correct answer

A. 1.5 ~J

Step-by-step solution

x _ i =(50-25) cm =25 ~cm x _ f =(60-25) cm =35 ~cm So, work done = U aligned & = 1 2 ~K [ x _ f ^2- x _ i ^2 ] & = 1 2 50 [35^2-25^2 ] 10⁻⁴ & =1.5 ~J aligned

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