# M3: Elastic stringsWatch

Announcements
#1
so, I think the LHS is wrong. I think EPE(A)+ KE(A) are right, but not sure about work. so, work done in the direction of movement + the kinetic energy in direction of movement + EPE in direction of movement?

the EPE(A) is the energy stored in the string and KE(A) is it moving towards B?

when the particle has reached b, the KE = 0, and the work done = 0 because its at rest? so, is the EPA the energy stored in the string as a result of its movement to A, and its ability to move back up the slope as its gone past its natural extension due to projection?
0
1 year ago
#2
(Original post by Maths&amp;physics)
so, I think the LHS is wrong. I think EPE(A)+ KE(A) are right, but not sure about work. so, work so in the direction of movement + the kinetic energy in direction of movement + EPE in direction of movement (as object was in equilibrium)?

the EPE(A) is the energy stored in the string and KE(A) is it moving to B.

when the particle has reached b, the KE = 0, because its at rest. so, is the EPA the energy stored in the string as a result of its movement to A, and its ability to move back up the slope as its gone past its natural extension due to projection?
The particle moves down the slope, so it loses gpe and kinetic energy as it come to instantaneous rest. The string gains elastic potential energy.

So loss of GPE+loss of KE=Final EPE-Initial EPE.

U can form an equation and solve for V let me know if further clarification is needed.

I think if i understand ur notation correctly:

EPE(A)+KE(A)+GPE(A)=EPE(B)+KE(B) +GPE(B)
0
#3
(Original post by Shaanv)
The particle moves down the slope, so it loses gpe and kinetic energy as it come to instantaneous rest. The string gains elastic potential energy.

So loss of GPE+loss of KE=Final EPE-Initial EPE.

U can form an equation and solve for V let me know if further clarification is needed.

I think if i understand ur notation correctly:

EPE(A)+KE(A)+GPE(A)=EPE(B)+KE(B) +GPE(B)
I did work done (force times . distance) and not GPE. 0
#4
(Original post by Shaanv)
The particle moves down the slope, so it loses gpe and kinetic energy as it come to instantaneous rest. The string gains elastic potential energy.

So loss of GPE+loss of KE=Final EPE-Initial EPE.

U can form an equation and solve for V let me know if further clarification is needed.

I think if i understand ur notation correctly:

EPE(A)+KE(A)+GPE(A)=EPE(B)+KE(B) +GPE(B) so I get: which is wrong.

aslo, why did you use GPE and not work done? thanks
0
#5
(Original post by Notnek)
...
(Original post by RDKGames)
.....
have you guys seen this question? thanks
0
1 year ago
#6
(Original post by Maths&physics)
have you guys seen this question? thanks
I have, but I'm doing my own maths so I can't help right now.
0
#7
(Original post by RDKGames)
I have, but I'm doing my own maths so I can't help right now.
ok, thanks 0
1 year ago
#8
(Original post by Maths&amp;physics)
so, I think the LHS is wrong. I think EPE(A)+ KE(A) are right, but not sure about work. so, work done in the direction of movement + the kinetic energy in direction of movement + EPE in direction of movement?

the EPE(A) is the energy stored in the string and KE(A) is it moving towards B?

when the particle has reached b, the KE = 0, and the work done = 0 because its at rest? so, is the EPA the energy stored in the string as a result of its movement to A, and its ability to move back up the slope as its gone past its natural extension due to projection?
I get v=root(ga/2).

I used GPE as the particle moves down the slope and its vertical height decreases therefore it must lose kinetic energy.

If u were trying to say why i didn’t use Work Done Against fricition, thats because the plane is smooth.
0
#9
(Original post by Shaanv)
I get v=root(ga/2).

I used GPE as the particle moves down the slope and its vertical height decreases therefore it must lose kinetic energy.

If u were trying to say why i didn’t use Work Done Against fricition, thats because the plane is smooth.
I see where I went wrong in the calculation.

ok, that makes sense. what is the equation you got for GPE? thanks
0
1 year ago
#10
(Original post by Maths&amp;physics)
I see where I went wrong in the calculation.

ok, that makes sense. what is the equation you got for GPE? thanks
GPE lost=(mga)/4.

I just used mgh. Where h=(a*sin(30))/2. If u have a good diagram u should see how i arrived at this answer
0
1 year ago
#11
(Original post by Maths&amp;physics)
thats how I got there, was yours the same????
I drew the triangle with hypotenuse a/2 as that was the further distance travelled and proceeded to work out the length of what u have labelled h.
0
#12
(Original post by Shaanv)
I drew the triangle with hypotenuse a/2 as that was the further distance travelled and proceeded to work out the length of what u have labelled h.
sorry, the diagram was a little inaccurate but thats how I got it. so, its the same as yours I'm sure. O (height) = (1a/2)sin(30)
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Bournemouth University
Wed, 19 Feb '20
• Buckinghamshire New University
Wed, 19 Feb '20
• University of Warwick
Thu, 20 Feb '20

### Poll

Join the discussion

Yes (47)
31.54%
No (76)
51.01%
Don't know (26)
17.45%