A force of 100N stretches an elastic string to a total length of 20cm. If an additional force of 100N stretches the string 5cm further, Find the natural length of the string

Zimaimoisilf

2 years ago

I don't understand my school how did you arrive at that answer

EmicksG

3 years ago

Note that the change in the value of x should be 40 because of the extra 100N added

Kpcofgs

1 year ago

my school, i dont understand how did you get the 5 in your solution

Truss

1 year ago

We will use Hooke’s Law, which states that the extension of an elastic string is proportional to the applied force , i.e.,

F = kx

where is the stiffness (spring constant) of the string.

Step 1: Find the Spring Constant

Given: A force of 100N stretches the string to 20cm (from its natural length).

Another 100N force stretches it 5cm more.


From Hooke’s Law:

100 = kx_1

where is the initial extension.
From the additional force:

100 = k(5)

Solving for :

k = \frac{100}{5} = 20 \text{ N/cm}

Step 2: Find the Initial Extension

Using the first force equation:

100 = 20 x_1

x_1 = \frac{100}{20} = 5 \text{ cm}

Step 3: Determine the Natural Length

The total stretched length is 20 cm.

The total extension is cm.

Natural length = Total length – Extension


20 - 10 = 10 \text{ cm}

Answer:

C. 10 cm