Question

An ideal spring of negligible mass is 11.00cm long when nothing is attached to it. When you hang a 3.05-kg weight from it, you measure its length to be 12.40cm .

If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.
Express your answer numerically. If there is more than one answer, enter each answer, separated by a comma.
=

Answer 1

The force constant of the spring is 5.45 N/m and the unloaded length of the spring is 0.200 m.

Step-by-step explanation:

The force constant of a spring can be determined using Hooke's law, which states that the force exerted by a spring is directly proportional to its displacement. In this case, we can use the equation F = kx, where F is the force, k is the force constant, and x is the displacement of the spring. To find the force constant, we can use the lengths and masses given in the question.

(a) To find the force constant, we can use Hooke's law with the given lengths and masses. For the 0.300-kg mass, the displacement is 0.550 m (0.750 m - 0.200 m). So, F = k * 0.550 m. Solving for k, we get k = F / 0.550 m. Substituting the values, k = (0.300 kg * 9.8 m/s^2) / 0.550 m. Using a calculator, we get k = 5.45 N/m.

(b) The unloaded length of the spring refers to its length when no mass is attached. In this case, the length of the spring is 0.200 m when no mass hangs from it.

Answer 2

0.2067m or 0.2067m

Explanation;

Let lenght of spring= Lo= 11cm=0.110m

It is hang from a mass of

3.05-kg having a length of L1= 12.40cm= 0.124m

Force required to stretch the spring= Fkx

But weight of mass mg= kx then K= Mg/x

K= 3.05-kg× 9.8)/(0.124m-.110m)

K=2135N

But potential Energy U= 0.5Kx

X=√ 2U/k

√(2*10)/2135

X=0.0967m

The required new length= L2= L0 ±x

=

.110m ± 0.0967m

X= 0.2067m or 0.2067m hence the total lenghth