Question

A spring of spring constant 1000N/m is initially stretched by 2 cm from natural length. The external work required to stretch it further by another 4 cm is

Answer 1

Answer:

1 n - m

Explanation:  

Given A spring of spring constant 1000 N/m is initially stretched by 2 cm from natural length. The external work required to stretch it further by another 4 cm is

Given A spring of spring constant 1000 N/m is initially stretched by 2 cm from natural length. The external work required to stretch it further by another 4 cm is

We know that W = 1/2  (x 2^2 – x 1^2)

                           W = 1/2 x K (x2 ^2 – x1 ^2 )

                           W =  1/2  x 1000( (6 / 100)^2 – (2 / 100)^2)

                           W = 500 (6/100 + 2/100)(6/100 – 2/100) ( because a^2 – b^2 = (a + b)(a – b) )

                           W = 500 (8/100)(4/ 100)

                           W = 16 / 10

                          W = 1.6 J or 1.6 N – m

Answer 2

A spring force is directly proportional to displacement of spring from its mean position or stretching of spring from natural length.

e.g., $F\propto x$

or, F = Kx , where k is spring constant e.g., 1000 N/m

so, F = 1000x

now, workdone is required to stretch 2cm or 0.02m to (2cm + 4cm) = 6cm or 0.06m from its natural length , W = $\int\limits^{0.06}_{0.02}{1000x}\,dx$

= $1000\left[\frac{x^2}{2}\right]^{0.06}_{0.02}$

= $1000\frac{(0.06)^2-(0.02)^2}{2}$

= 500( 36 × 10^-4 - 4 × 10^-4) J

= 500 × 32 × 10^-4 J

= 160 × 10^-2 J

= 1.6 J