How Does a Mass and Spring System Behave on an Inclined Plane?

[besenji]

Homework Statement

A spring (80 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 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. (Intro 1 figure)


If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest?

If the mass is attached to the spring, how far up the slope will the mass move before coming to rest?

Now the incline has a coefficient of kinetic friction uk. 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 uk?

1/2mv^2+mgh=1/2mv^2+mgh
1/2kx^2

This problem has been giving me a ton of trouble.

I am sure that it involves the conservation of energy theorm, ad the spring constant.

I just need help with he correct equations to be set in the right direction.

 

[alphysicist]

Homework Statement

A spring (80 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 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. (Intro 1 figure)


If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest?

If the mass is attached to the spring, how far up the slope will the mass move before coming to rest?

Now the incline has a coefficient of kinetic friction uk. 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 uk?

1/2mv^2+mgh=1/2mv^2+mgh

This energy equation (for conservation of energy) does not have the spring potential energy in it. It's just an extra term for each side:

1/2mv^2+mgh + 1/2 k x^2=1/2mv^2+mgh + 1/2 k x^2

where the terms on the left are for the initial point, and the terms on the right are for the final point. When you apply this equation to the first question, what does this equation become? What do you get as the answer?


(For the third question, energy is not conserved. How will the equation change for that?)

 

[thomate1]

I would approach this problem by first identifying all the given information and variables. We have a spring with a known spring constant (80 N/m) and equilibrium length (1.00 m), a compressed length (0.50 m), a mass (2.2 kg), and an inclined slope (41 degrees). We are also given two scenarios: one where the mass is not attached to the spring, and one where it is attached.

To solve for the distance the mass will move before coming to rest, we need to use the conservation of energy principle. In both scenarios, the initial energy of the system is stored in the compressed spring, and the final energy is all in the form of potential energy due to the mass being lifted up the slope.

For the first scenario, where the mass is not attached, the potential energy gained by the mass will be equal to the potential energy lost by the spring. We can set up the equation as follows:

1/2kx^2 = mgh

Where k is the spring constant, x is the compressed length, m is the mass, g is the acceleration due to gravity, and h is the vertical distance the mass will move up the slope. Solving for h, we get:

h = (1/2kx^2)/mg

Plugging in the given values, we get h = 0.125 m.

For the second scenario, where the mass is attached, the potential energy gained by the mass will be equal to the potential energy lost by the spring, plus the work done by the spring on the mass. We can set up the equation as follows:

1/2kx^2 + 1/2mv^2 = mgh

Where v is the final velocity of the mass. We can use the conservation of energy principle again to find the value of v. The initial energy of the system (stored in the compressed spring) will be equal to the final energy of the system (all in the form of kinetic energy of the mass). We can set up the equation as follows:

1/2kx^2 = 1/2mv^2

Solving for v, we get:

v = √(kx^2/m)

Now, we can plug this value of v into our original equation and solve for h:

1/2kx^2 + 1/2m(kx^2/m) = mgh