You find that if you hang a 1.25kg weight from a vertical spring, it stretches 3.75cm.

a) What is the force constant of this spring?

b) How much mass should hang from this spring so it will stretch 8.13 cm from its original unstretched length?

(a) 326.67 N/m

(b) 2.71 kg

Explanation:

(a)

From hook's Law,

F = ke ... Equation 1

Where F = force, k = spring constant, e = extension.

Note: Here the force on the spring is the weight of the mass

W = ke

Where W = weight of the mass

But,

W = mg ... Equation 2

where m = mass and g = acceleration due to gravity.

substituting equation 2 into equation 1

mg = ke ... Equation 3

make k the subject of the equation

k = mg/e ... Equation 4

Given: m = 1.25 kg, g = 9.8 m/s², e = 3.75 cm = 0.0375 m

Substitute into equation 4

k = 1.25*9.8/0.0375

k = 326.67 N/m

(b)

making m the subject of equation 3

m = ke/g ... Equation 5

Given: e = 8.13 cm = 0.0813 m, k = 326.67 N/m

Substitute into equation 5

m = 326.67*0.0813/9.8

m = 2.71 kg.

(A) k = 327N/m

(B) m = 2.71kg

Explanation:

The weight of the 1.25kg is the force acting on the spring. W = weight = mg = F = kx

That is

mg = kx

Therefore k = mg/x

Given m = 1.25kg and x = 3.75cm = 0.0375m

k = 1.25 * 9.8 / 0.0375 = 327N/m

(B) from the relation mg = kx

Given x = 8.13cm = 0.0813m.

m = kx / g = 327 * 0.0813 / 9.8 = 2.17kg.