Question

A 25.0 [cm] long spring stands vertically on the ground, with its lower end secured in a base. A brick with mass, 1.5 [kg] is held 40 [cm] directly above the spring and dropped onto the spring. The spring compresses to a length of 17.0 [cm] before starting to launch the brick back upward. What is the spring’s spring constant?

Answer 1

compression in spring = 25-17 = 8

change in potential energy of brick when it joins the spring = mg*(40 + 8)

which is equal to the stored potential energy in sprint = 1/2*k*(8^2)

so k = 3/2mg = 3/2*1.5*10 = 2250 N/m

Step-by-step explanation:

Answer 2

1837.5 N/m

Step-by-step explanation:

From the law of conservation of energy,

The potential energy of the brick = energy stored in the spring

mgh = 1/2ke²................. Equation 1

Where m = mass of the brick, h = height, g = acceleration due to gravity, k = spring constant, e = extension

Make k the subject of the equation

k = 2mgh/e².................. Equation 2

Given: m = 1.5 kg, e = 25-17 = 8 cm = 0.08 m, h = 40 cm = 0.4 m, g = 9.8 m/s²

Substitute into equation 2

k = 2(1.5)(9.8)(0.4)/(0.08²)

k = 1837.5 N/m

Hence the spring constant of the spring = 1837.5 N/m