Recent content by Serella_Madole

Thanks for the correction. I fixed my mistake.

You should get -Kx2/2 +Fx = mv2/2

Hello, sorry for the late response Hello, sorry for the late response! I do understand what each of these variables stand for. And I am familiar with integration.

Sorry, I did not see the previous message until after I had posted my response.

v = -kx^2 + Fext (x) / m

-F(spring) +F(ext) = ma -F(spring) +F(ext) = m dv/dt integrate -F(spring)x + F(ext)x = mv v= -fspringx +f(ext)x/m

-F spring exerts + Fexternal = 0 because afterwards they no longer exert a force on the object.

Opps! The force the spring exerts on the can is leftward which I chose to be negative because the spring is trying to return back to it's on. The external force is in the rightward direction which I chose to be positive.

Okay, I did draw the diagram as Mister T suggested. I noticed the change in position, because of the force applied the initial conditions (x =0) do not work... So I would need to solve for the new x which will become the x initial, which I believe that is what you, rmthomps, was hinting at...

Eventually, the spring will need to return back to the equilibrium point where x = 0. Otherwise the object will be deformed and won't be able to return back to it's original state.

I think that it is 1/2kx^2. But because it is stretched would it be negative?

I do not think it is at maximum stretch at .4m. I think it can go further. But I am not sure as how to proceed with this.

Homework Statement One end of a horizontal spring with force constant (76.0 N/m) is attached to a vertical post. A 4.00-kg can of beans is attached to the other end. The spring is initially neither stretched nor compressed. A constant horizontal force of 57.0 N is then applied to the can, in...

Sorry for the late response! Here is all the information accompanying the question! Part A If frictional forces do −1.05×104 J of work on her as she descends, how fast is she going at the bottom of the slope? Take free fall acceleration to be g = 9.80 m/s2. v = 32.1 m/s Part B Now moving...

I had to determine the final velocity on my own which is 17.8 m/s. The initial velocity is zero based on the first part of the question. "A 61.0-kg skier starts from rest at the top of a ski slope of height 70.0 m."