Find K: Hooke's Law & Massless Spring Homework

[Physics197]

Homework Statement

A massless spring of length 0.260 m is in its relaxed position. It is compressed to 68.0 percent of its relaxed length, and a mass M=0.240 kg is placed on top and released from rest (shown on the right). The mass then travels vertically, taking 1.60 s to reach the top of its trajectory. Calculate the spring constant.

Homework Equations

F = -kx
Spring potential = 0.5kx^2

The Attempt at a Solution

I would prefer if someone explained the concept behind it so that I can work it out myself, rather than people just posting answers.

 

[rock.freak667]

So if the spring is compressed to 68% of its initial length, to what length is is compressed to?

Thus by how much is it compressed by?

Now they say that, it takes 1.6s to reach the top of the trajectory. So all the energy stored in the spring is converted into what sort of energy?

When you figure out that, then can you make an equation relating the energy stored to the converted energy?


(Also remember, it is only being influenced by gravity g)

 

[kerimek]

As a scientist, it is important to understand the concepts and principles behind the equations and calculations we use. In this case, Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This can be represented by the equation F = -kx, where k is the spring constant and x is the displacement.

In order to find the spring constant, we need to use the information given in the problem. We know that the spring is initially compressed to 68.0% of its relaxed length, so the displacement x is equal to 0.68 times the relaxed length. We also know that the mass placed on the spring is 0.240 kg and that it takes 1.60 s for the mass to reach the top of its trajectory.

Using the equation for spring potential, we can set the potential energy at the top of the trajectory equal to the initial potential energy of the compressed spring. This can be represented as 0.5kx^2 = mg(0.68l), where m is the mass, g is the acceleration due to gravity, and l is the relaxed length of the spring.

Solving for k, we get k = (2mg)/(0.68l^2). Plugging in the values, we get k = (2*0.240 kg*9.8 m/s^2)/(0.68*0.260 m)^2 = 148.7 N/m.

Therefore, the spring constant for this massless spring is approximately 148.7 N/m. It is important to note that this value may vary slightly depending on the accuracy of the measurements and calculations used.