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eleastic potential energy watch

1. a 200g object is on the end of a spring constant of 2.5N/m. calculate the elastic potential energy stored in the spring?
2. (Original post by chrisyabantony14)
a 200g object is on the end of a spring constant of 2.5N/m. calculate the elastic potential energy stored in the spring?
Im gonna assume that one end of the spring is fixed and that the particle is hanging in equilibrium directly below the point of suspension.

Firstly i recommend drawing a diagram and labelling all the forces clearly. From there note that the particle is hanging in equilibrium. What does this tell you about the resultant force on the particle?

Now you should be in a position to find x the extension. Then its just a matter if plugging your x value into the correct formula for elastic potential energy.

Let me know if u didn’t get any bits.
3. (Original post by Shaanv)
Im gonna assume that one end of the spring is fixed and that the particle is hanging in equilibrium directly below the point of suspension.

Firstly i recommend drawing a diagram and labelling all the forces clearly. From there note that the particle is hanging in equilibrium. What does this tell you about the resultant force on the particle?

Now you should be in a position to find x the extension. Then its just a matter if plugging your x value into the correct formula for elastic potential energy.

Let me know if u didn’t get any bits.
don't want to be rude but can you exlain me that wiht some equation please
4. convert 200g into newtons- 10N=1kg
from this you can work out the extension by dividing 2N by 2.5N/m
=0.8 metres
EPE=1/2 ke^2
1/2 x 2.5 x 0.8^2
=0.8J
(I think)
5. (Original post by chrisyabantony14)
don't want to be rude but can you exlain me that wiht some equation please
T-0.2g=0
T=kx
Solve for x.
EPE=0.5kx^2
6. (Original post by Angel_Chen)
convert 200g into newtons- 10N=1kg
from this you can work out the extension by muliplying 2.5N/m by 2N
=5 metres
EPE=1/2 ke^2
1/2 x 2.5 x 5^2
=31.25
(I think)
This is wrong.
7. (Original post by Shaanv)
This is wrong.
F= kx
so 1.962/2.5
Right?
8. (Original post by Angel_Chen)
convert 200g into newtons- 10N=1kg
from this you can work out the extension by dividing 2N by 2.5N/m
=0.8 metres
EPE=1/2 ke^2
1/2 x 2.5 x 0.8^2
=0.8J
(I think)
9. (Original post by chrisyabantony14)
elastic potential energy
10. (Original post by Angel_Chen)
elastic potential energy
don't worry but thanks again but 200g is 0.2kg now what shou i do?
11. (Original post by Angel_Chen)
convert 200g into newtons- 10N=1kg
from this you can work out the extension by dividing 2N by 2.5N/m
=0.8 metres
EPE=1/2 ke^2
1/2 x 2.5 x 0.8^2
=0.8J
(I think)
Yep, you're right! I'm also getting it as 0.769j
12. (Original post by chrisyabantony14)
don't worry but thanks again but 200g is 0.2kg now what shou i do?
if 10 Newtons=1kg
0.2kg=...
13. (Original post by Iluv2edgedsword)
F= kx
so 1.962/2.5
Right?
why did u use 1.962 and not 2??? Doesn't 10 Newtons=1kg
14. (Original post by Angel_Chen)
if 10 Newtons=1kg
0.2kg=...
2N thanks again
15. (Original post by chrisyabantony14)
2N thanks again
Np :P
16. (Original post by Angel_Chen)
why did u use 1.962 and not 2??? Doesn't 10 Newtons=1kg
oh yeah !!! I quite forgot! In my gcses i used to use 10 N but now in my AS its 9.81... doesnt matter...
17. (Original post by Iluv2edgedsword)
oh yeah !!! I quite forgot! In my gcses i used to use 10 N but now in my AS its 9.81... doesnt matter...
thanx for the clarification :P
18. (Original post by Angel_Chen)
thanx for the clarification :P
Np
19. thanks to everyone of you for helping me
20. (Original post by chrisyabantony14)
a 200g object is on the end of a spring constant of 2.5N/m. calculate the elastic potential energy stored in the spring?
I got an 8 in my Physics mock eeee!!!!

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