# Find the initial speed of a block before it hits a spring

W =ΔKE

## The Attempt at a Solution

found work done by spring

-1.3 =ΔKE
-1.3 = KEFINAL - KEINITIAL
-1.3 = .5(0.25)(VFINAL2) - .5(0.25)(VINITIAL2)

Since v final is 0 since it stops momenteraly at the bottom of the spring
-1.3 = - .5(0.25)(VINITIAL2)
sqrt((-1.3/-.125)) = VINITIAL
VINITIAL = 3.22m/s

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tnich
Homework Helper

## Homework Statement

View attachment 224911
View attachment 224915

W =ΔKE

## The Attempt at a Solution

found work done by spring
View attachment 224913

-1.3 =ΔKE
-1.3 = KEFINAL - KEINITIAL
-1.3 = .5(0.25)(VFINAL2) - .5(0.25)(VINITIAL2)

Since v final is 0 since it stops momenteraly at the bottom of the spring
-1.3 = - .5(0.25)(VINITIAL2)
sqrt((-1.3/-.125)) = VINITIAL
VINITIAL = 3.22m/s
I assume you want to know why the answer to question 3 is incorrect.
I see you have answered the work question with a precision of 2 significant digits. Then you used it to calculate speed at a precision of 3 significant digits.

I don't think that is the problem here, i think i am doing something wrong with the calculation logic here

-1.26 =ΔKE
-1.26 = KEFINAL - KEINITIAL
-1.26 = .5(0.25)(VFINAL2) - .5(0.25)(VINITIAL2)

Since v final is 0 since it stops momenteraly at the bottom of the spring
-1.26 = - .5(0.25)(VINITIAL2)
sqrt((-1.26/-.125)) = VINITIAL
VINITIAL = 3.17m/s

using the unrounded answer, still incorrect

So i was using the value of work for the spring on the block to find the initial velocity of the block, i am not sure if that is correct

tnich
Homework Helper
-1.26 =ΔKE
-1.26 = KEFINAL - KEINITIAL
-1.26 = .5(0.25)(VFINAL2) - .5(0.25)(VINITIAL2)

Since v final is 0 since it stops momenteraly at the bottom of the spring
-1.26 = - .5(0.25)(VINITIAL2)
sqrt((-1.26/-.125)) = VINITIAL
VINITIAL = 3.17m/s

using the unrounded answer, still incorrect
Yes, but you have given too many significant digits in the answer.

isukatphysics69
tnich
Homework Helper
So i was using the value of work for the spring on the block to find the initial velocity of the block, i am not sure if that is correct
It looks right to me. I got the same answer independently.

isukatphysics69
It looks right to me. I got the same answer independently.
i just tried 3.2m/s incorrect

tnich
Homework Helper
i just tried 3.2m/s incorrect
I think we forgot about gravitational potential energy.

isukatphysics69
I think we forgot about gravitational potential energy.
my book says ΔKE = KFINAL - KINITIAL = WA+WG
since energy is conserved isn't WA = -WG

tnich
Homework Helper
my book says ΔKE = KFINAL - KINITIAL = WA+WG
since energy is conserved isn't WA = -WG
I don't know what WA and WG represent.
Total energy (kinetic + potential) is conserved. There is definitely a change in gravitational potential energy between the point at which the block hits the spring and the point at which it bottoms out.

isukatphysics69
I don't know what WA and WG represent.
Total energy (kinetic + potential) is conserved. There is definitely a change in gravitational potential energy between the point at which the block hits the spring and the point at which it bottoms out.
youre right

i remember from class last week prof was talking about potential energy.

tnich
Homework Helper
I don't know what WA and WG represent.
Total energy (kinetic + potential) is conserved. There is definitely a change in gravitational potential energy between the point at which the block hits the spring and the point at which it bottoms out.
So you need to add up the potential energy (due to gravity and the spring) and kinetic energy. Do this at both the point where the block hits the spring and where it bottoms out.
Also, think about the appropriate sign for the potential energy of the spring in this equation.

isukatphysics69
So then i think the gravity force has to be integrated from initial to bottom

wait i already have the work that gravity did while spring is compressed

it was 0.29J

omg i am an idiot

tnich
Homework Helper
it was 0.29J
OK, you have all of the pieces. Now you need to write your conservation of energy equation.

isukatphysics69
ok the answer is 2.8 m/s

tnich
Homework Helper
ok the answer is 2.8 m/s
Yes. That's what I got, too.

isukatphysics69
thank you i completely forgot about potential energy