Question

4. Mollie is using springs to hang art projects. (I'm not sure why. Ask her.) She hangs a 3.

on one particular spring and it stretches from 30 cm to 45 cm long.
a) What is the spring constant for that spring?
b) If she hangs a 2 kg picture beneath the first one, how long will the spring get?

Answer 1

Explanation:524 Hz

Answer 2

answers

part a = 196 N/m

part b = 55 cm or 0.55 m

Step-by-step explanation

we can solve this using Hooke's law where spring force is equal to the product of displacement and spring constant

`F_s =kx`

part a

rearrange Hooke's law to solve for k (spring constant)

`F_s =kx\\k = (F_s)/(x)`

the spring force caused by the painting here is equal to the weight of the painting

`F_s = mg\\k = (mg)/(x)`

m = 3 kg

g = 9.8 m/s^2

x = 45 - 30 = 15 cm = 0.15m

`k = (3*9.8)/(0.15)\\k = 196 (N)/(m)`

part b

since we now know the spring constant, we can find the displacement caused by a 2kg picture by rearranging Hooke's law to solve for x, where spring force is still equal to weight

`F_s =kx\\x = (F_s)/(k)\\x = (mg)/(k)`

m = 2 kg

g = 9.8 m/s^2

k = 196 N/m

`x = (2*9.8)/(196) \\x = 0.1m`

the spring is stretched an additional 0.1 meters or 0.1 * 100 = 10 centimeters

45 cm + 10 cm = 55 cm or 0.55 m