Find the distance the spring compresses

[joemama69]

Homework Statement

a block of mass m slides along ahrizontal table with speed v. At s = 0, it hits a spring with spting constat k and begins to experience a friction oforce. the coefficient of frictio is u, find the distance the spring compresses when it momentairly comes to rest.

Homework Equations

The Attempt at a Solution

i believe the friction part is umgs, is this correct

.5mv² = .5ks² + ugms

if this is correct, how do i get the s alone

 

[Doc Al]

i believe the friction part is umgs, is this correct

.5mv² = .5ks² + ugms

if this is correct, how do i get the s alone

It's a quadratic equation. Put it in standard form and solve it.

(That attachment doesn't seem to match the problem you described.)

 

[joemama69]

s = -mg + or - (((mg)² - 4(.5k)(mgh))/k)^.5

 

[Doc Al]

s = -mg + or - (((mg)² - 4(.5k)(mgh))/k)^.5

How did you get this from the equation you gave in your first post?

I suspect you're solving a different problem.

 

[joemama69]

haha you i had a very similar one sorry about that


s = -umg + or minus (((umg)² - 4(.5k)(-.5mv²))/k)^.5

 

[Doc Al]

s = -umg + or minus (((umg)² - 4(.5k)(-.5mv²))/k)^.5

That's almost right. (Redo it more carefully. Each term must have the same units.)

Also, only one of the + or - choices will make sense for this problem.

 

[joemama69]

i redid and came up with the same

.5ks² + umgs - .5mv² = 0

a = .5k
b = umg
c = -.5mv²

-b - \sqrt{b^2-4ac}/2a

-umg - \sqrt{()^2 - 4(.5k)(-.5mv^2)}/2(.5k)

s = -umg - \sqrt{()^2 +kmv^2}/k

 

[Doc Al]

-b - \sqrt{b^2-4ac}/2a

Careful with how you do the division by 2a. Everything is divided by 2a, not just the square root term. (Use parentheses to show this.)

And don't forget that it's + or -, not just -.

 

[joemama69]

ok

i think it should be minus because the spring is being compressed, do you concure

 

[Doc Al]

i think it should be minus because the spring is being compressed, do you concure

No...

 

[joemama69]

s = -umg + (((umg)2 - 4(.5k)(-.5mv2))^.5/k

s = -umg + (((umg)2 + kmv²)^.5/k

 

[Doc Al]

s = -umg + (((umg)2 - 4(.5k)(-.5mv2))^.5/k

s = -umg + (((umg)2 + kmv²)^.5/k

You still haven't corrected the "division by 2a" problem from post #8.

 

[joemama69]

= (-umg + ((umg)2 + kmv²)^.5)/k

the answer how ever is Wnc = -m(v0)^2/2 (1 + k/bmg)^-1 after simplifying