#### To determine

**To find:**

The work done in stretching the spring from 12cm to 20cm.

#### Answer

3.2 J

#### Explanation

**1) Concept:**

i. Use Hooke’s Law

ii. Use the formula for work done in moving an object from a to b

**2) Theorem:**

Hooke’s Law states that the force required to maintain a spring stretched x units beyond its natural length is proportional to x.

fx=kx

where, k is a positive constant called the spring constant.

**3) Given:**

force is 30N

**4) Formula:**

Work done in moving an object from a to b

∫abfxdx

**5) Calculations:**

Using Hooke’s Law,

fx=kx

Substituting the values,

30N=k15cm-12cm

Convert centimeter to meter.

Therefore,

30=k0.15-0.12

30=0.03k

Dividing by 0.03,

k=1000

Hence, fx=kx=1000x

Natural length of the spring is 12cm.

So when it is stretched from 12cm to 20cm, it is stretched by 8cm beyond its natural length.

Therefore, by using the formula for work done,

Work needed to stretch the spring 0.08m beyond its natural length =∫00.08fxdx

Substituting the value of fx,

W=∫00.081000x dx

After integrating,

W=1000x220.080

Applying the Fundamental Theorem of Calculus,

W=5000.08-02=5000.0064=3.2

**Conclusion:**

The work done in stretching the spring from 12cm to 20cm is 3.2 J.