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

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

Answer 2

Answer:

1.6 Joules

Explanation:

Work done by a spring = ½kx²

k = the spring constant

x = the extension

In this case we have x1 and x2. The work done in stretching the spring will therefore be :

Work = ½k(x1² - x2²)

x1 = 2 + 4 = 6cm

In meters = 6/100 = 0.06 m

x2 = 2 cm

In meters = 2/100 = 0.02 m

We substitute these value in the formula above.

W = 1/2 × 1000 × (0.06² - 0.02²)

W = 500 × 0.0032 = 1.6 Joules

The work done in stretching the spring is therefore 1.6 Joules.