Question

4. A spring of unknown force constant is attached to a rod so that it hangs vertically. With nomass attached, the spring has a length of 12.0 cm. When a student attaches a mass of 50 grams,the spring stretches to a length of 18.0 cm. Determine the force constant for this spring.

Answer 1

Answer:

The force constant is 8.33N/m

Explanation:

Using Hooke's law to solve the problem which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the force constant or elastic constant

e is the extension of the material

Given mass = 50g = 0.05kg

F = mg = 0.05×10

F = 0.5N

If the initial length of the spring is Lo = 12cm

Final length L = 18cm(after load has been applied)

extension caused by the applied load on the spring = L-Lo

extension = 18cm-12cm

extension = 6cm = 0.06m

Substituting this values in the formula F = ke to get the force constant k, we will have;

k = F/e

k = 0.5N/0.06m

k = 8.33N/m

The force constant is 8.33N/m

Answer 2

Answer:

8.17 N/m

Explanation:

From Hook's law,

F = ke................. Equation 1

Where F = Force on the spring, k = force constant of the spring, e = extension.

From the question,

The weight attached to the spring is the force acting on it.

W = mg = F

Where W = weight, m = mass, g = acceleration due to gravity.

Therefore,

mg = ke

make k the subject of the equation

k = mg/e................... Equation 2

Given: m = 50 g = 0.05 kg, e = 18-12 = 6 cm = 0.06 m.

Constant: g = 9.8 m/s.

Substitute into equation 2

k = 0.05(9.8)/0.06

k = 8.17 N/m

Hence the force constant for the spring = 8.17 N/m