Did i do this problem right? (Work by Spring and KE)

[Lori]

Homework Statement

Homework Equations

W= KE
Wspring = .5kx^2

The Attempt at a Solution

since W = .5mv^2 = Wspring = .5x^2
i solve for v and get that v = sqrt(kx^2/m)
if i plug in multiply the x by 2, the 2 squares and i get 4. the square root of 4 is just 2 and it comes out as a constant , and so velocity is 2 times faster

Did i go through the problem correctly?[/B]

 

[PeroK]

Homework Statement

View attachment 214618

Homework Equations

W= KE
Wspring = .5kx^2

The Attempt at a Solution

since W = .5mv^2 = Wspring = .5x^2
i solve for v and get that v = sqrt(kx^2/m)
if i plug in multiply the x by 2, the 2 squares and i get 4. the square root of 4 is just 2 and it comes out as a constant , and so velocity is 2 times faster

Did i go through the problem correctly?[/B]

You got the right answer for the right reasons. You might like to think, however, about how you could organise your working better.

 

[Abhishek kumar]

Homework Statement

View attachment 214618

Homework Equations

W= KE
Wspring = .5kx^2

The Attempt at a Solution

since W = .5mv^2 = Wspring = .5x^2
i solve for v and get that v = sqrt(kx^2/m)
if i plug in multiply the x by 2, the 2 squares and i get 4. the square root of 4 is just 2 and it comes out as a constant , and so velocity is 2 times faster

Did i go through the problem correctly?[/B]

You are right but if you have to write what is relation between heights

 

[Lori]

You are right but if you have to write what is relation between heights

mgh + 1/2mv^2=PE of Spring + mgh
mgh + 1/2mv^2 = .5kx^2 + 0
mgh + .5mv^2 = .5k(2x)^2
mgh + .5mv^2 =4 (.5kx^2)

since m and g are constant in the left side of the equation, the coefficient 4 will affect the height and the speed of the object and the final KE
height will be 4 times as big and the velocity would be twice as fast at final, while the final KE will times 4

 

[Abhishek kumar]

mgh + 1/2mv^2=PE of Spring + mgh
mgh + 1/2mv^2 = .5kx^2 + 0
mgh + .5mv^2 = .5k(2x)^2
mgh + .5mv^2 =4 (.5kx^2)
height will be 4 times as big and the velocity would be twice as fast at final

When block will reach its maximum height velocity will be zero at that instant simply applying energy conservation at that instant you will get the result

 

[Lori]

When block will reach its maximum height velocity will be zero at that instant simply applying energy conservation at that instant you will get the result

How would you write the equation?

How is the one I wrote above incorrect?

In energy conservation E2 = E1

(Final energy) PE2 + KE2 = PE + KE1 (initial energy)

 

[Abhishek kumar]

How would you write the equation?

How is the one I wrote above incorrect?

In energy conservation E2 = E1

(Final energy) PE2 + KE2 = PE + KE1 (initial energy)

When block reach max height at that instant velocity seized and no kinetic energy
. 5kx^2=mgh

 

[haruspex]

seized

Ceased.

 

[Abhishek kumar]

Ceased.

Thank you for correcting me

 

[Lori]

Thank you for correcting me

Oh I see. So in this spring problem. The initial energy comes from PE of spring.

The final energy comes from the gravitational energy since kinetic energy is 0 (because the object stops)

 

[Abhishek kumar]

Oh I see. So in this spring problem. The initial energy comes from PE of spring.

The final energy comes from the gravitational energy since kinetic energy is 0 (because the object stops)[

Oh I see. So in this spring problem. The initial energy comes from PE of spring.

The final energy comes from the gravitational energy since kinetic energy is 0 (because the object stops)

Yes you got it