Law of Conservation of Energy with a Spring

[adrenalinefix]

Homework Statement

A spring has its right end fixed and is installed on a horizontal table so that the free end, in equilibrium, is at x = 3.00 m. A 1.65 kg block coming from the left slides along the table. When it passes the origin, it is moving at 5.58 m/s. It strikes the spring, compresses it momentarily, and is then sent back toward the left, where it eventually comes to rest at the point x = 1.50 m. The coefficient of kinetic friction between the block and the table is 0.300. By what distance was the spring compressed?

Homework Equations

KE=1/2mv^2
PEspr.=1/2kx^2
Wfric.= (Uk)(Fn)d

The Attempt at a Solution

I'm just not sure what to do. It should be so easy, with only KE of the block and PE of the spring, along with the Work done by friction. I don't really see how it's possible to solve for a displacement/compression of the spring when the spring constant isn't known.

 

[tiny-tim]

Welcome to PF!

Hi adrenalinefix ! Welcome to PF!

(have a mu : µ and try using the X² tag just above the Reply box )

A spring has its right end fixed and is installed on a horizontal table so that the free end, in equilibrium, is at x = 3.00 m. A 1.65 kg block coming from the left slides along the table. When it passes the origin, it is moving at 5.58 m/s. It strikes the spring, compresses it momentarily, and is then sent back toward the left, where it eventually comes to rest at the point x = 1.50 m. The coefficient of kinetic friction between the block and the table is 0.300. By what distance was the spring compressed?
…
I'm just not sure what to do. It should be so easy, with only KE of the block and PE of the spring, along with the Work done by friction. I don't really see how it's possible to solve for a displacement/compression of the spring when the spring constant isn't known.

Hint: why would the spring constant matter? How would it affect the work done?

 

[adrenalinefix]

I really appreciate your help but I'm really dense. Please let me see if my interpretation of the question is correct. So a block moving to the right is first retarded by friction. It then strikes a spring, thereby compressing this spring a given distance. From there the spring releases and propels the object at a certain speed. This object then slows down due to the force of friction once more.

Assuming this is correct, then you can first calculate the final kinetic energy given an initial velocity used to calculate the initial kinetic energy, and the work done by friction. This is -Wfr= delta(KE). From here, all of the kinetic energy should be converted to PE of the spring. Then all of this PE of the spring should be converted to KE for the block.

Based on my reasoning, it's pointless to solve for anything later than the KE after the release of the spring, as they are superfluous. But if nothing more were to be calculated, then the answer can't be determined, suggesting that my original logic is off.

Based on what you said about work, I wasn't saying that a different K value would change how much work was there I was just saying that given the equation 1/2Kx^2, which has two unknowns, I can't solve for one variable if I don't have the other(and certainly K is the one that should be given since x is the final variable to be solved).Please post any additional hints. thanks for the help

 

[tiny-tim]

Assuming this is correct, then you can first calculate the final kinetic energy given an initial velocity used to calculate the initial kinetic energy, and the work done by friction. This is -Wfr= delta(KE). From here, all of the kinetic energy should be converted to PE of the spring. Then all of this PE of the spring should be converted to KE for the block.

ok, stop there!

KE is converted to PE, and then all the PE is converted back to KE …

so in your equation W = energy_initial - energy_final, you don't need to know what the PE was.

The only unknown is x, and so one equation should be enough to find it.