Question

A spring has natural length 23 cm. Compare the work W1 done in stretching the spring from 23 cm to 33 cm with the work W2 done in stretching it from 33 to 43 cm. (Use k for the spring constant) W

Answer 1

Answer:

The relation between $W_{1} \ and \ W_{2}$ is $W_{2} = 3 \ W_{1}$

Step-by-step explanation:

Natural length = 0.23 m

Spring stretches from 23 cm to 33 cm. now

Work done $W_{1}$ in stretching the spring

$W_{1} = \int\limits^a_b {kx} \, dx$

where b = 0 & a = 0.1 m

$W_{1} = k [\frac{x^{2} }{2} ]$

With limits b = 0 & a = 0.1 m

Put the values of limits we get

$W_{1} = k [\frac{0.1^{2} }{2} ]$

$W_{1} = 0.005 k$ ------- (1)

Now the work done in stretching the spring from 33 cm to 43 cm.

$W_{1} = \int\limits^a_b {kx} \, dx$

With limits b = 0.1 m to a = 0.2 m

$W_{2} = k [\frac{x^{2} }{2} ]$

With limits b = 0.1 m to a = 0.2 m

$W_{2} = k [\frac{0.2^{2} - 0.1^{2} }{2} ]$

$W_{2} =0.015$

$\frac{W_{2} }{W_{1} } = \frac{0.015}{0.005}$

$\frac{W_{2} }{W_{1} } =3$

Thus

$W_{2} = 3 \ W_{1}$

This is the relation between $W_{1} \ and \ W_{2}$.