Zenarthra
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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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laur21
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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 ?
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Zenarthra
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(Original post by 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 ?
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.
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laur21
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(Original post by 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.
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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Zenarthra
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(Original post by 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.
Here is what they have put:
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Davelittle
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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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laur21
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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.
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Zenarthra
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(Original post by 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
I do agree but i just wanted to understand the application of work energy principle to this situation. ;D
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Zenarthra
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#9
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(Original post by 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.
its ok, thank for your help but i have figured it out now.
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