If a wire 30 cm long is extended to 30.5 cm by a force of 300 N. find the strain energy of wire
Answer Details
Energy stored in a stretched wire (strain energy) = average force x extension = (1)/2 Fe = (1)/2 x 300 N (30.5 - 30.0) x 10-2 m = (300)/2 x 0.5 x 10-2 = 0.75J