Question

A spring of force constant 'k' is stretched by a small length 'x'. Find the work done in stretching it further by a small length 'y'.​

Answer 1

Answer:

Initial stretching of the spring is x.

Initial potential energy U

i

=

2

1

kx

2

Final stretching of the spring is x+y.

Final potential energy of the spring U

f

=

2

1

k(x+y)

2

=

2

1

kx

2

+

2

1

ky

2

+

2

1

k(2xy)

Work done W=U

f

−U

i

=

2

1

kx

2

+

2

1

ky

2

+

2

1

k(2xy)−

2

1

kx

2

⟹ W=

2

ky

(2x+y)

Answer 2

$\Large\bf {\orange {\underline { \underline {Solutíon}} : - }}$

Let $\sf \: W_{1}$ is the work done in stretching a spring of froce constant 'K' through a length "X" Then,

$\sf W_{1} = \frac{1}{2} {Kx}^{2}$ .........(1)

Let $\sf \: W_{2}$ is the work done in stretching the spring a length (x + y) .Then

$\sf W_{2} = \frac{1}{2} K {(x + y)}^{2}$ .........(2)

$\therefore$ Additional work done, to increase the elongation by "y" is

$\sf W = W_{2} - W_{1}$ ⠀⠀⠀⠀[using (1) and (2).]

$\sf W = \frac{1}{2} K {(x + y)}^{2} - \frac{1}{2} K {x}^{2}$

$\therefore$ $\boxed{ \sf W = \frac{1}{2} Ky(y + 2x)}$