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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 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 ?
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 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.
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 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.
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 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
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 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.
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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