A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5cm to 15cm is,
16 J
8 J
32 J
24 J
work done in extending the spring is W = 12 K(x22 − x12) ∴given x2 = 15×10−2m and x1 = 5×10−2m force constant k=800N/m substitute given values in above equation of work W= 12 (800){(15x10 -2)2 − (5x10-2)2} work done =8 joule