#### To determine

**To explain:**

How are *W2* and *W1* related when work *W1* done in stretching the spring from *20 cm* to *30 cm* is compared to work *W2* done in stretching the spring from *30 cm* to *40 cm*.

#### Answer

*W2=3W1*

#### Explanation

**1) Concept:**

Use Hook’s law, then use the formula for work done to find *W1* and *W2* and compare them.

**2) Law and Formula:**

Hook’s law:

The force required to stretch a spring *x* units beyond its natural length is proportional to *x,*

*fx=kx* Where *k* is a positive constant called the spring constant.

Work done:

*W=∫abf(x)dx*

**3) Calculation:**

From the Hook’s law,

*fx=kx* where *k* is a positive constant called the spring constant.

When the spring is stretched from *20 cm* to *30 cm*, the amount stretched is *10cm=0.1m*

From the formula, the work *W1* done in stretching the spring from *20 cm* to *30 cm* is

*W1=∫00.1kxdx*

Simplify,

*W1=kx2200.1*

*=k0.122-[k022]*

*=0.01k2*

*W1=1200k*

When the spring is stretched from *30 cm* to *40 cm*, the amount stretched is *10cm=0.1m*. And the stretch is increases from 0.1 to 0.2.Hence, from the formula the work *W2* done in stretching the spring from *30 cm* to *40 cm* is

*W2=∫0.10.2kxdx*

Simplify,

*W1=kx220.10.2*

*=k0.222-[k0.122]*

*=0.04k2-0.01k2*

*=0.032k*

*W2=3200k*

*W2=3 ·1200k*

*W2=3W1*

Therefore, *W2=3W1*

**Conclusion:**

The relation is

*W2=3W1*