JEE Main Physics question
An elastic string of unstretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is
Answer: (D)
Solution
W = PE(x+y) − PE(x) = ½k(x+y)² − ½kx² = ½k[(x+y)²−x²] = ½k[2xy+y²] = ½ky(2x+y).