Need Help on an Elastic Potential Energy Problem on an incline

[portuwhat]

Homework Statement

A spring (k=75 N/m) has an equilibrium length of 1 m. The spring is compressed to a length of .5 m and a mass of 2 kg is placed at its free end on a frictionless slope which makes an angle of 41 degrees with respect to the horizontal. The spring is then released. a) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? b) If the mass is attached to the spring, how far up the slope will the mass move before coming to rest? c) Now the incline has a coefficient of kinetic friction. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction?

Homework Equations

Conservation of mechanical energy
.5mvi^2 + mgyi + .5 kxi^2 = .5 mvf^2 + mgyf + .5 kxf^2

The Attempt at a Solution

So I began using the conservation of mechanical energy but I'm not even sure where to begin with that. What do I say the y is or x is in the equation? I have no clue, is there anyway someone can explain this to me without giving me the answer? Thanks.

 

[tiny-tim]

Welcome to PF!

Conservation of mechanical energy
.5mvi^2 + mgyi + .5 kxi^2 = .5 mvf^2 + mgyf + .5 kxf^2
…
What do I say the y is or x is in the equation?

Hi portuwhat! Welcome to PF!

This is just geometry …

y is the height (measured vertically, of course).

x is the extended length of the spring (measured along the 41º slope).

Have a go!

 

[baba89]

As a scientist, it's important to approach problems like this using a systematic and logical approach. First, let's identify the key information given in the problem: the spring constant (k), equilibrium length (x), compressed length (xi), mass (m), slope angle (θ), and coefficient of kinetic friction (μ). These values can be used to calculate the potential energy stored in the spring and the gravitational potential energy of the mass at different points along the slope.

To begin, we can use the given values to calculate the potential energy stored in the spring at the compressed length (xi). This can be done using the equation for elastic potential energy: U = ½ kΔx^2. In this case, Δx is the change in length from the equilibrium length to the compressed length, which is 0.5 m. So, U = ½ (75 N/m)(0.5 m)^2 = 9.375 J.

Next, we can use this value to calculate the initial kinetic energy of the mass as it is released from the compressed spring. We know that the mass will initially have zero velocity, so the initial kinetic energy will be zero.

Now, we can use the conservation of mechanical energy equation to find the final velocity of the mass as it reaches the equilibrium position of the spring. This can be done by setting the initial and final energies equal to each other and solving for the final velocity (vf). Once we have the final velocity, we can use it to calculate the final height (yf) of the mass using the equation for gravitational potential energy: U = mgy.

For part a, where the mass is not attached to the spring, we can use the final height (yf) to calculate the distance the mass will move up the slope before coming to rest. This can be done using basic trigonometry and the slope angle (θ).

For part b, where the mass is attached to the spring, we need to also consider the potential energy stored in the spring at the equilibrium position (xf) and the change in height (Δy) as the mass moves up the slope. This can be done using the same approach as part a, but now we have an additional energy term to consider in the conservation of mechanical energy equation.

For part c, we need to use the given information about the coefficient of kinetic friction (μ) to find the final velocity of the mass at the equilibrium position. This can