JAMB - Physics (2010 - No. 6)
If a wire 30 cm long is extended to 30.5 cm by a force of 300 N. find the strain energy of wire
0.75 J
7.50 J
75.00 J
750.00 J
Explanation
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
= 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
Comments (0)
