Calculating Motion of a Mass Attached to a Spring on an Inclined Plane

[emc92]

A spring (75 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.

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

b) Now the incline has a coefficient of kinetic friction μ. 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 μ?

 

[emc92]

i know that if the mass is NOT attached to the spring, the formula to find d (distance up the slope) is 1/2kx^2=mgh. i don't understand how attaching the two together will change it..
and to answer part b, i need the equation to answer part a, right?

 

[gneill]

i know that if the mass is NOT attached to the spring, the formula to find d (distance up the slope) is 1/2kx^2=mgh. i don't understand how attaching the two together will change it..

A spring that's not attached to its load will impart KE to the load until the load's speed exceeds the spring's speed. This usually occurs at the equilibrium point of the spring. At this point the mass is "launched" and continues upwards using its own KE while the spring slows and then recoils. The mass travels as high as its KE will carry it against gravity.

If the mass is attached to the spring, the spring gets pulled past the equilibrium point and begins to drag the mass back -- the mass' KE is now fighting both the stretching spring and gravity.

and to answer part b, i need the equation to answer part a, right?

The added complication of track friction is going to modify the energy conservation equation.

 

[emc92]

A spring that's not attached to its load will impart KE to the load until the load's speed exceeds the spring's speed. This usually occurs at the equilibrium point of the spring. At this point the mass is "launched" and continues upwards using its own KE while the spring slows and then recoils. The mass travels as high as its KE will carry it against gravity.

If the mass is attached to the spring, the spring gets pulled past the equilibrium point and begins to drag the mass back -- the mass' KE is now fighting both the stretching spring and gravity.


The added complication of track friction is going to modify the energy conservation equation.

so for part a, do i need to add 1/2kx^2 to the mgh side to account for the spring being stretched?

 

[gneill]

so for part a, do i need to add 1/2kx^2 to the mgh side to account for the spring being stretched?

Well, you need to account for the initial PE stored in the spring in some fashion so that your conservation equation will agree with the "locations" of the energy throughout the motion.

I would consider setting the origin at the starting position of the mass on the compressed spring so that the initial gravitational PE is zero along with the KE, and then giving the spring its initial energy by doing something clever with the PE equation for the spring. The total energy of the system should start out equal to the PE stored in the spring.

The conservation of energy equation involves the sum of all the energies at any given time, and it should equal a constant value. In this case,

$PE_s + PE_g + KE = E_o$

Write expressions for each of the PE terms of the left that reflect their values with respect to position (for the spring PE, think offset from equilibrium).

 

[dianathom]

Well, you need to account for the initial PE stored in the spring in some fashion so that your conservation equation will agree with the "locations" of the energy throughout the motion.

I would consider setting the origin at the starting position of the mass on the compressed spring so that the initial gravitational PE is zero along with the KE, and then giving the spring its initial energy by doing something clever with the PE equation for the spring. The total energy of the system should start out equal to the PE stored in the spring.

The conservation of energy equation involves the sum of all the energies at any given time, and it should equal a constant value. In this case,

$PE_s + PE_g + KE = E_o$

Write expressions for each of the PE terms of the left that reflect their values with respect to position (for the spring PE, think offset from equilibrium).

Many thanks for your inputs! I found them extremely helpful! :)