Question

launch task:

1. spring constant: (use hookes law \\(f = -kx\\) to determine the k spring constant for the following)

a) 12n weight with spring extension of 2.7cm
b) 28n weight with spring extension of 5cm

1. using your answer from (a) determine the energy stored in the spring using \\(e_p = \frac{1}{2}kx^2\\)
2. calculate the springs extension (x value) if now a 3.6kg mass was attached. (you will need to find fw)

Explanation:

Convert units to SI standards

To solve these physics problems accurately, we must first convert all given values into standard SI units.

• Force/Weight is already in Newtons (\(\text{N}\)).
• Extensions must be converted from centimeters (\(\text{cm}\)) to meters (\(\text{m}\)) by dividing by \(100\):
• For 1(a): \(x = 2.7\text{ cm} = 0.027\text{ m}\)
• For 1(b): \(x = 5\text{ cm} = 0.05\text{ m}\)
• Mass in question 3 is given in kilograms (\(\text{kg}\)), which is already the standard SI unit: \(m = 3.6\text{ kg}\).

Calculate spring constants for Question 1

Using Hooke's Law, the magnitude of the restoring force exerted by a spring is related to its extension by:

$$F = kx \implies k = \frac{F}{x}$$

• For 1(a):
• Given: \(F = 12\text{ N}\), \(x = 0.027\text{ m}\)
• Calculation:

$$k = \frac{12\text{ N}}{0.027\text{ m}} \approx 444.44\text{ N/m}$$

• For 1(b):
• Given: \(F = 28\text{ N}\), \(x = 0.05\text{ m}\)
• Calculation:

$$k = \frac{28\text{ N}}{0.05\text{ m}} = 560\text{ N/m}$$

Calculate elastic potential energy for Question 2

Using the spring constant \(k \approx 444.44\text{ N/m}\) from 1(a), we calculate the elastic potential energy (\(E_p\)) stored in the spring when extended by \(x = 0.027\text{ m}\):

$$E_p = \frac{1}{2}kx^2$$

• Calculation:

$$E_p = \frac{1}{2} \times 444.44\text{ N/m} \times (0.027\text{ m})^2$$

$$E_p = 222.22 \times 0.000729 \approx 0.162\text{ J}$$

Calculate spring extension for Question 3

First, find the weight force (\(F_w\)) acting on the spring due to the \(3.6\text{ kg}\) mass. Using gravity \(g \approx 9.8\text{ m/s}^2\) (or \(9.81\text{ m/s}^2\)):

$$F_w = m \cdot g = 3.6\text{ kg} \times 9.8\text{ m/s}^2 = 35.28\text{ N}$$

(If using \(g = 9.81\text{ m/s}^2\), \(F_w = 35.316\text{ N}\))

Next, assuming we use the spring from 1(a) with \(k \approx 444.44\text{ N/m}\), we solve for the new extension \(x\):

$$F_w = kx \implies x = \frac{F_w}{k}$$

• Calculation:

$$x = \frac{35.28\text{ N}}{444.44\text{ N/m}} \approx 0.0794\text{ m} = 7.94\text{ cm}$$

Answer:

No.	Problem	Answer
1(b)	Spring constant for \(28\text{ N}\) weight with \(5\text{ cm}\) extension	\(k = 560\text{ N/m}\)
2	Energy stored in spring 1(a) using \(E_p = \frac{1}{2}kx^2\)	\(E_p \approx 0.162\text{ J}\)
3	Extension \(x\) for a \(3.6\text{ kg}\) mass attached to spring 1(a)	\(x \approx 0.079\text{ m}\) (or \(7.9\text{ cm}\))