Where will the end of the spring line up with the ruler marks?

[chazgurl4life]

A spring with k = 55 N/m hangs vertically next to a ruler. The end of the spring is next to the 15 cm mark on the ruler. If a 1.9 kg mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?

For this porblem i used the following :

W= delta Ke + delta PR
-Force of friction= 0+ ( Pe1-Pe0) + 1/2 kx^2
-Ukmgx= MG ( Y1-Y0) + 1/2 KX^2
SO WHEN I SUBSITUTE :

-Uk(1.9 KG)(9.8M/S^2)=(1.9)(9.8) +.5(55 N/M)x^2

But now i have two variables instead of one.

now when using this equation to solve I am become confused because there isn't a coefficient given. I just wanted to know if i could use any other equations

 

[cepheid]

W= delta Ke + delta PR
-Force of friction

Well, this equation doesn't make a lot of sense. If W is the work done, and delta Ke is the change in kinetic energy, than these two are equal, by the work energy theorem. What is delta PR?

Why have you included friction? What is there in this system that you think would cause a significant amount of friction?

I think that energy analysis may not be the best way to go. There is probably another method of solving the problem. Hint: what condition must be satisfied for the mass to stop moving?

 

[chazgurl4life]

im sorry delta Pr= Delta Pe ...just a typo

my professor suggested we use the above equation to solve the problem

for the mass to stop moving: shouldn't the force of the string stop from the mass moving vertically along the ruler.

 

[topsquark]

im sorry delta Pr= Delta Pe ...just a typo

my professor suggested we use the above equation to solve the problem

for the mass to stop moving: shouldn't the force of the string stop from the mass moving vertically along the ruler.

I'm confused about two things. First, where the heck did the friction come from? If the mass is falling on the end of the spring, it isn't rubbing up against anything to create friction. Second, what precisely are you solving for when you say "where will the end of the spring line up with the ruler marks?" I would normally say we are looking for where the mass ends up when it's stationary, but the equation you are using seems to imply you are letting the mass drop and are looking for the lowest point in the oscillation.

Last thing:

-Force of friction= 0+ ( Pe1-Pe0) + 1/2 kx^2

You seem to be equating a force with an energy. Careful!

-Dan

 

[plusaf]

A spring with k = 55 N/m hangs vertically next to a ruler. The end of the spring is next to the 15 cm mark on the ruler. If a 1.9 kg mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?

For this porblem i used the following :

W= delta Ke + delta PR
-Force of friction= 0+ ( Pe1-Pe0) + 1/2 kx^2
-Ukmgx= MG ( Y1-Y0) + 1/2 KX^2
SO WHEN I SUBSITUTE :

-Uk(1.9 KG)(9.8M/S^2)=(1.9)(9.8) +.5(55 N/M)x^2

But now i have two variables instead of one.

now when using this equation to solve I am become confused because there isn't a coefficient given. I just wanted to know if i could use any other equations

um, I'm a little confused.
you know the coeffecient for the spring, which says that if you pull on it with a force of 55N, it will stretch 1m.

you hang a 1.9kg mass on the end of the spring.
it stretches to a new length.

can you figure out how many N of force the 1.9kg mass exerts on the end of the spring?

the spring coefficient essentially tells how far it will stretch for a given force on it. in the Earth's gravitational field, how many Newtons of force will the 1.9kg mass pull on the end of the spring.

f=kx?

did i miss something??