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A spring is 30 cm long and extends by 5 cm for each 100 gm hung on it? What is the springs length when 300gm is hung from it?what is the springs length z when n 100 gm masses are hung from it? What is the springs length when n =4?

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Hello! Let's solve the problem.

The elongation of the spring is directly proportional to the applied force, according to Hooke's Law. The formula for this relationship is given by:

\[ F = k \cdot x \]

where:
- \( F \) is the force applied to the spring,
- \( k \) is the spring constant (a measure of the stiffness of the spring), and
- \( x \) is the displacement or elongation of the spring.

In this case, the spring elongates by 5 cm for each 100 gm (0.1 kg) hung on it. Therefore, the spring constant \( k \) can be calculated as:

\[ k = \frac{x}{F} \]

Substitute \( x = 0.05 \) meters (5 cm converted to meters) and \( F = 0.1 \) kg into the formula to find \( k \).

Once you have the spring constant \( k \), you can use it to find the elongation \( z \) for any given force \( F \) using the formula:

\[ z = \frac{F}{k} \]

Now, for the specific question:
1. For 300 gm (0.3 kg) hung on the spring, substitute \( F = 0.3 \) kg into the \( z \) formula to find the spring's length.
2. For \( n \) masses of 100 gm each, the total force \( F \) is \( 0.1n \) kg. Substitute this into the \( z \) formula.
3. For \( n = 4 \), substitute \( n = 4 \) into the \( z \) formula to find the spring's length.
maryoomsaggaf

Answer :

1. Extension for 300 gm: The spring's length becomes 45 cm.

2. Extension for n 100 gm masses: For n masses, the extension is 5n cm.

3. Length when n = 4: With 4 masses of 100 gm each, the spring's length is 50 cm.

Step-by-step explanation:

To solve this problem, we can use the concept of Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it, as long as the elastic limit of the spring is not exceeded.

Given:

- Original length of the spring (unstretched): 30 cm

- Extension per 100 gm: 5 cm

1.  Extension for 300 gm:

  - Each 100 gm adds 5 cm of extension.

  - For 300 gm, which is 3 times 100 gm, the extension would be 3 x 15 cm.

  - Therefore, the new length of the spring with 300 gm hung from it would be 30 + 15 = 45 cm.

2. Extension for n 100 gm masses:

  - Since each 100 gm adds 5 cm of extension, for 'n' 100 gm masses, the extension would be 5 x n cm.

  - If 'n' represents the number of 100 gm masses, then the total extension would be 5n cm.

3. Length when n = 4:

  - Given n = 4 (four 100 gm masses), the extension would be 5 x 4 = 20 cm.

  - Therefore, the new length of the spring would be 30 + 20 = 50 cm when 4 masses of 100 gm each are hung from it.