Question

A mass is attached to a spring is 0.2 kg. The spring's constant is 40 N/m. The mass is stretched to a position of 0.3 m and released. The final position is 0.1 m. #1) Find the change in potential energy of the system. #2) What is the work done by the spring?

#1) Change in P.E.=(1/2)(40 N/m)(0.3 m)^2
" "= 1.8 Joules
#2)W=[1/2(40N/m).3m^2]-[1/2(40N/m).1m^2]
W= 1.6 Joules
Can you please tell me if I am correct. I thank you in advance for your time. JL

Answer 1

I saw this but do not have the picture.
Is this vertical or horizontal?

Is the equilibrium position .1 metert of stretch?

Final when? Does it not oscillate?

Change of U when you do what?

Work done by spring during what motion?

Answer 2

I'm sorry, I wasn't sure how to post the picture. The picture I have is horizontal and stretched in the positive direction. The position when the spring is unstrained is stretched to Xo .3m. Then released the spring is at Xf .1m. Stretch to the right. Thanks Damon for responding, I wasn't sure why. I will do better when asking to check my work.

Answer 3

I wasn't sure why I wasn't getting any responses on this, but now I know. I will do better when asking for help next time. Thank you Damon, I really appreciate your response and telling me why I wasn't getting any help.

Answer 4

If it is horizontal and you pull it .3 m to the right then the potential energy is (1/2) k(.3)^2
= (1/2)(40)(.09)
= 1.8 Joules

The work done by the spring is indeed the change in potential energy in going from .3 m back to .1 meters stretch (thee work appears as kietic energy in the motion of the mass)
1.8 Joules - (1/2)(40)(.1)^2
1.8 - .2 = 1.6 Joules

so I think you did fine.