Question

An ideal spring of negligible mass is 12.00 cm long when nothing is attached to it. When you hang a 3.55 kg  object from it, you measure its length to be 13.40 cm.

Part A

If you wanted to store 10.0 J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.

Express your answer in centimeters to three significant figures. If there is more than one answer, separate them by a comma.

Answer 1

here,

the original length , l = 12 cm = 0.12 m

when

the hanging mass , m = 3.55 kg

the length , l1 = 13.4 cm = 0.134 m

let the spring constant of the spring be K

K * (l1 - l) = m * g

K * ( 0.134 - 0.12) = 3.55 * 9.81

K = 2487.5 N/m

let the total length of spring be l2 when the energy stored is 10 J

PE = 0.5 * K * ( l2 - l)^2

10 = 0.5 * 2487.5 * ( l2 - 0.12)^2

solving for l2

l2 = 0.210 m = 21.1 cm or

l2 = 0.0303 m = 3.03 cm

the new length of the spring is 21.0 cm or 3.03 cm