Question

When a spring of force constant \(200~\text{Nm}^{-1}\) is compressed by a force of \(10~\text{N}\), then the compression in the spring is

• A. \(5~\text{m}\)
• B. \(5~\text{cm}\)
• C. \(0.5~\text{m}\)
• D. \(0.5~\text{cm}\)

Hint:

Use Hooke’s Law: \(x = \frac{F}{k}\) — Always convert to desired units (meters to cm if needed).

Answer 1

The Correct Option is B

Hooke’s Law relates the force \(F\) applied on a spring to its compression \(x\) through the spring constant \(k\): \[ F = kx \Rightarrow x = \frac{F}{k} \] Given: \[ F = 10~\text{N}, \quad k = 200~\text{Nm}^{-1} \Rightarrow x = \frac{10}{200} = 0.05~\text{m} = 5~\text{cm} \] So, the compression in the spring is 5 cm.

Answer 2

Given Data:
- Force constant (spring constant), \( k = 200~\text{Nm}^{-1} \)
- Force applied on the spring, \( F = 10~\text{N} \)

What is asked?
- The compression \( x \) in the spring when this force is applied.

Concept:
According to Hooke's law, the restoring force exerted by a spring is directly proportional to the displacement (compression or extension) from its natural length:
\[ F = kx \] where,
- \( F \) is the force applied,
- \( k \) is the spring constant,
- \( x \) is the compression (or extension) of the spring.

Calculations:
Rearranging the formula to find \( x \):
\[ x = \frac{F}{k} \] Substitute the given values:
\[ x = \frac{10~\text{N}}{200~\text{Nm}^{-1}} = 0.05~\text{m} \] Convert meters to centimeters:
\[ 0.05~\text{m} = 0.05 \times 100 = 5~\text{cm} \] Interpretation:
This means the spring compresses by 5 cm under the applied force of 10 N.

Final answer:
\[ \boxed{5~\text{cm}} \]