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
- A1.5 ~J
- B2 ~J
- C3.5 ~J
- 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