If a load of 9kg is suspended on a wire, the increase in length is 4.5 mm. The force constant of the wire is

The correct answer is 1.96 x 10⁴ N / m.

To find the force constant of the wire, we can use Hooke's Law, which states that the force applied is directly proportional to the extension (increase in length) produced:

$F=k·\mathrm{\Delta x}$

where:
• $F$ is the applied force (equal to the weight of the load suspended on the wire),
• $k$ is the force constant of the wire, and
• $\mathrm{\Delta x}$ is the extension of the wire.

Rearranging the formula to solve for the force constant $k$ gives:

$k=\frac{F}{\mathrm{\Delta x}}$

Let's identify the given values and convert them into standard SI units:

• Mass of the suspended load, $m=9\text{kg}$
• Acceleration due to gravity, $g=9.8{\text{m/s}}^2$
• Applied force (weight), $F=m·g=9·9.8=88.2\text{N}$
• Increase in length, $\mathrm{\Delta x}=4.5\text{mm}=4.5·{10}^{-3}\text{m}$

Now, substitute the values of force and extension into the formula for $k$:

$k=\frac{88.2}{4.5·{10}^{-3}}$

$k=19.6·{10}^3\text{N/m}$

$k=1.96·{10}^4\text{N/m}$

Therefore, the force constant of the wire is 1.96 x 10⁴ N / m.