# M3: Elastic strings Watch

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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 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?

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#2

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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?

**Maths&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?

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)

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(Original post by

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)

**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)

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**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

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(Original post by

...

**Notnek**)...

(Original post by

.....

**RDKGames**).....

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#6

(Original post by

have you guys seen this question? thanks

**Maths&physics**)have you guys seen this question? thanks

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(Original post by

I have, but I'm doing my own maths so I can't help right now.

**RDKGames**)I have, but I'm doing my own maths so I can't help right now.

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#8

(Original post by

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?

**Maths&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 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.

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(Original post by

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.

**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.

ok, that makes sense. what is the equation you got for GPE? thanks

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#10

(Original post by

I see where I went wrong in the calculation.

ok, that makes sense. what is the equation you got for GPE? thanks

**Maths&physics**)I see where I went wrong in the calculation.

ok, that makes sense. what is the equation you got for GPE? thanks

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

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#11

(Original post by

thats how I got there, was yours the same????

**Maths&physics**)thats how I got there, was yours the same????

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(Original post by

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.

**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.

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