Question

A 60-kilogram person stands on a compression spring. The spring constant is 2500 Newtons per meter What is the displacement of the spring?

A)0.00392M
B) 0.024m
C)0.1176m
D)0.2352M

Answer 1

The answer is 0.024m

60-kilogram divided by 2500 Newtons per meter or 60/2500 is 0.024

Answer 2

The displacement
`(\( \Delta x \))` of the compression spring when a 60-kilogram person stands on it, with a spring constant of 2500 N/m, is approximately -0.2352 meters. The negative sign indicates the direction of the displacement. Therefore, the correct answer is D.

To find the displacement
`(\( \Delta x \))` of the spring, you can use Hooke's Law:

`\[ F = -k \cdot \Delta x \]`

Where:

- F is the force applied (weight in this case),

- k is the spring constant, and

-
`\( \Delta x \)` is the displacement of the spring.

Rearrange the formula to solve for
`\( \Delta x \)`:

`\[ \Delta x = -(F)/(k) \]`

Substitute the given values:

`\[ \Delta x = -(mg)/(k) \]\[ \Delta x = -\frac{(60 \, \text{kg})(9.8 \, \text{m/s}^2)}{2500 \, \text{N/m}} \]`

Calculate the result:

`\[ \Delta x \approx -0.2352 \, \text{m} \]`

The negative sign indicates that the displacement is in the opposite direction to the applied force. Therefore, the correct answer is:

D)
`\( -0.2352 \, \text{m} \)`