# M2

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#1

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!
0
6 years ago
#2

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 ?
0
#3
(Original post by laur21)

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.
0
6 years ago
#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.
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.
0
#5
(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:
0
6 years ago
#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
0
6 years ago
#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.
0
#8
(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
0
#9
(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.
0
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