# M2

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For the last part, if i consider work energy principle:

Why isn't it Total change in energy = Total workdone

3500 + 2800 = 0.5(10)(2^2) - 0.5(10)(v^2) ?

Thanks!

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#2

The answer is v~11.57

and... the workdone of Friction force is always negative, this force is against the move of the particle.

Also, you should know that is kinetic energy from final form - kinetic energy from initial form.

Ecf-Eci=Total workdone

1/2*m*(v^2) - 1/2*m*(vA)^2=F*d-Ff*d

where va is the speed of the particle in A point.

Edit: you can also solve this using Galiei formula: v2^2=v1^2+2aD

where

v1 is the initial speed

v2 is the final speed

D-distance

and a is acceleration

btw,is that from A-levels ?

and... the workdone of Friction force is always negative, this force is against the move of the particle.

Also, you should know that is kinetic energy from final form - kinetic energy from initial form.

Ecf-Eci=Total workdone

1/2*m*(v^2) - 1/2*m*(vA)^2=F*d-Ff*d

where va is the speed of the particle in A point.

Edit: you can also solve this using Galiei formula: v2^2=v1^2+2aD

where

v1 is the initial speed

v2 is the final speed

D-distance

and a is acceleration

btw,is that from A-levels ?

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reply

(Original post by

The answer is v~11.57

and... the workdone of Friction force is always negative, this force is against the move of the particle.

Also, you should know that is kinetic energy from final form - kinetic energy from initial form.

Ecf-Eci=Total workdone

1/2*m*(v^2) - 1/2*m*(vA)^2=F*d-Ff*d

where va is the speed of the particle in A point.

Edit: you can also solve this using Galiei formula: v2^2=v1^2+2aD

where

v1 is the initial speed

v2 is the final speed

D-distance

and a is acceleration

btw,is that from A-levels ?

**laur21**)The answer is v~11.57

and... the workdone of Friction force is always negative, this force is against the move of the particle.

Also, you should know that is kinetic energy from final form - kinetic energy from initial form.

Ecf-Eci=Total workdone

1/2*m*(v^2) - 1/2*m*(vA)^2=F*d-Ff*d

where va is the speed of the particle in A point.

Edit: you can also solve this using Galiei formula: v2^2=v1^2+2aD

where

v1 is the initial speed

v2 is the final speed

D-distance

and a is acceleration

btw,is that from A-levels ?

Why isnt it Total workdone on particle = Total change in energy?

And yes its A level M2.

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#4

(Original post by

They said the only answer is 12ms^-1 :/

Why isnt it Total workdone on particle = Total change in energy?

And yes its A level M2.

**Zenarthra**)They said the only answer is 12ms^-1 :/

Why isnt it Total workdone on particle = Total change in energy?

And yes its A level M2.

You should know two type of forces:

conservative forces-they don't depend on path,only initial and final position.(gravity force,elastic force,etc)

non-conservative forces=dpend on path(Friction force,etc).

Let's say you have a particle in two positions A=initial and B=final

If you have ONLY conservative forces :

EA=EB where E is the energy from that position.

If you have at least one non-conservative forces you must use:

EB-EA=Workdone of non conservative forces where E is the energy from that position.

or

Ec=Kinetic energy

EcB-EcA=Workdone of resultant force.

---------------------------------------------------

You're right,but you don't have potential energy because the particle only moves on horizontally.So,that is 0.also you don't have elastic energy because there is no resort.

Mhm..sorry,I can't explain well because I have never used english related to physics.

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(Original post by

Well..I believe(hope) they want an integer,I don't think i have made any mistake for sure.

You should know two type of forces:

conservative forces-they don't depend on path,only initial and final position.(gravity force,elastic force,etc)

non-conservative forces=dpend on path(Friction force,etc).

Let's say you have a particle in two positions A=initial and B=final

If you have ONLY conservative forces :

EA=EB where E is the energy from that position.

If you have at least one non-conservative forces you must use:

EB-EA=Workdone of non conservative forces where E is the energy from that position.

or

Ec=Kinetic energy

EcB-EcA=Workdone of resultant force.

---------------------------------------------------

You're right,but you don't have potential energy because the particle only moves on horizontally.So,that is 0.also you don't have elastic energy because there is no resort.

Mhm..sorry,I can't explain well because I have never used english related to physics.

**laur21**)Well..I believe(hope) they want an integer,I don't think i have made any mistake for sure.

You should know two type of forces:

conservative forces-they don't depend on path,only initial and final position.(gravity force,elastic force,etc)

non-conservative forces=dpend on path(Friction force,etc).

Let's say you have a particle in two positions A=initial and B=final

If you have ONLY conservative forces :

EA=EB where E is the energy from that position.

If you have at least one non-conservative forces you must use:

EB-EA=Workdone of non conservative forces where E is the energy from that position.

or

Ec=Kinetic energy

EcB-EcA=Workdone of resultant force.

---------------------------------------------------

You're right,but you don't have potential energy because the particle only moves on horizontally.So,that is 0.also you don't have elastic energy because there is no resort.

Mhm..sorry,I can't explain well because I have never used english related to physics.

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#6

Haven't done M2 in ages put is the newton's second law method not a lot easier?

Just find the resultant force and divide it by 10 to get the acceleration then use v^2=u^2+2as to get the velocity at B

Just find the resultant force and divide it by 10 to get the acceleration then use v^2=u^2+2as to get the velocity at B

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#7

I got 11.57 because I have approximated a with 1.3,not 1.4 as they did and therefore there is a little difference.

Could you tell me where you don't understand,maybe i will be able to help.

Could you tell me where you don't understand,maybe i will be able to help.

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(Original post by

Haven't done M2 in ages put is the newton's second law method not a lot easier?

Just find the resultant force and divide it by 10 to get the acceleration then use v^2=u^2+2as to get the velocity at B

**Davelittle**)Haven't done M2 in ages put is the newton's second law method not a lot easier?

Just find the resultant force and divide it by 10 to get the acceleration then use v^2=u^2+2as to get the velocity at B

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(Original post by

I got 11.57 because I have approximated a with 1.3,not 1.4 as they did and therefore there is a little difference.

Could you tell me where you don't understand,maybe i will be able to help.

**laur21**)I got 11.57 because I have approximated a with 1.3,not 1.4 as they did and therefore there is a little difference.

Could you tell me where you don't understand,maybe i will be able to help.

0

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