# Find the distance the spring compresses

## 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 frictin oforce. the coefficient of frictio is u, find the distance the spring compresses when it momentairly comes to rest.

## The Attempt at a Solution

i belive the frictin part is umgs, is this correct

.5mv2 = .5ks2 + ugms

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

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Doc Al
Mentor

i belive the frictin 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.)

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

Doc Al
Mentor

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.

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

Doc Al
Mentor

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.

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
Mentor

-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 -.

ok

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

Doc Al
Mentor

i think it should be minus because the spring is being compressed, do you concure
No...

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

s = -umg + (((umg)2 + kmv2).5/k

Doc Al
Mentor

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.

= (-umg + ((umg)2 + kmv2).5)/k

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