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Mechanics M1 Help Needed =]

Hey can anyone help me with this question. i'll write it out as the book does and the topic we are doing is "dynamics of a particle in a straight line or plane":

a stone slides in a straight line across a frozen pond. given that the initial speed is 5ms(-1) and it slides 20m before coming to rest, calculate the coefficient of friction between the stone and the surface of the frozen pond.


we have used the formulas F = ma
and i know

v = 0
u = 5
s = 20
t = 8
a = 0.625

i'm just stuck of what to do now!!

help me please.
Thankyou.
Reply 1
i failed this module :p: (teacher left halfway thru the course...and all the rest of it.... lol)
Reply 2
thirdrate
That was unhelpful blahhh.

Next use F=ma

F is the friction force in this case and it's equal to

the coefficient of friction x the normal reaction

So

c.o.f. x normal reaction = m x a


i was actually going to post something like tht...but couldnt find the mu symbol for the coefficient of friction
Reply 3
thankyouuu everyone =]
Reply 4
thirdrate
\mu gives μ\mu


ahh thanks! :smile:
Reply 5
just wondering, is the coefficient 0.625/9.81?
Reply 6
noo may be 0.625 but not 9.81 is the acceleration of particle down wards ....
Reply 7
OCC++
just wondering, is the coefficient 0.625/9.81?


Yes, that's correct. Because the stones perpendicular forces are in equilibrium, R=mgR = mg; and replacing the R into Friction (uR), you'll get μmg\mu mg

F=ma[br][br][br]μmg=ma[br][br]μ=a/gF = ma[br][br][br]\mu mg = ma[br][br]\mu = a/g
Reply 8
v2=u2+2asv^2=u^2+2as
0=52+40a0=5^2+40a
a=2540a=\frac{25}{40}
F/m=2540F/m=\frac{25}{40}

F=μRF=\mu R
R=mgR=mg
F=mgμF=mg \mu

Therefore, 2540m=mgμ\frac{25}{40}m=mg \mu
μ=2540/g\mu=\frac{25}{40} /g

=0.064 or so
Reply 9
I can understand all the workings but why do you make Fmax equal to the Force? It is moving so friction is obviously max but does it necessarily have to equal the force?
Reply 10
the stone eventually comes to a stop so we know that the horizontal forces are now equal. Therefore Fmax is equal to the force.
Reply 11
Original post by spyer21
the stone eventually comes to a stop so we know that the horizontal forces are now equal. Therefore Fmax is equal to the force.


This thread is over 6 years old. I'm sure somehow during that time they've managed to answer the question
Reply 12
Original post by KaylaB
This thread is over 6 years old. I'm sure somehow during that time they've managed to answer the question

Yeah I know, I just thought that someone else who might be looking at this thread in the future may have the same question so I thought i might as well awnser it.

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