#### To determine

**To find:**

The natural length of the spring L

#### Answer

8 cm

#### Explanation

**1) Concept:**

i) Use Hook’s law and formula of work done for finding the natural length of the spring.

ii) 100cm=1m

**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=∫abfx dx

**3) Calculation:**

Let L be the natural length of the spring in meters.

6 J of work is needed to stretch a spring from 10 cm to 12 cm

That is 6 J of work is needed to stretch a spring from 0.1 m to 0.12 m

Means for W=6 J the length of spring L is stretched from 0.10-Lm to 0.12-Lm

By using the formula for work done

6=∫0.10-L0.12-Lkx dx=12kx20.10-L0.12-L=12k0.12-L2-0.10-L2

6=12k0.12-L2-0.10-L2

6=k20.0144-20.12L+L2-0.0100+20.10L-L2

6=k20.0144-0.24L-0.0100+0.20L

6=k20.0044-0.04L

12=k0.0044-0.04L …(1)

Similarly,

10 J of work is needed to stretch a spring from 12 cm to 14 cm

That is 10 J of work is needed to stretch a spring from 0.12 m to 0.14 m

Means for W=10 J the length of spring L is stretched from 0.12-Lm to 0.14-Lm

By using the formula for work done

10=∫0.12-L0.14-Lkx dx=12kx20.12-L0.14-L=12k0.14-L2-0.12-L2

10=12k0.14-L2-0.12-L2

10=k20.142-20.14L+L2-0.122-20.12L+L2

10=k20.0196-0.28L+L2-0.0144-0.24L+L2

10=k20.0196-0.28L+L2-0.0144+0.24L-L2

10=k20.0052-0.04L

20=k0.0052-0.04L …(2)

Subtracting the first equation from the second

20=0.0052k-0.04kL

__12=0.0044k-0.04kL__

8=0.0008k

Therefore,

k=10,000

Substitute k in equation (2)

20=(10,000)0.0052-0.04L

20=52-400L

Therefore,

L=32400 m=0.08 m

By concept ii),

L=8 cm

**Conclusion:**

The natural length of the spring is 8 cm