Mechanics M1 dynamics of a particle. Question help.?

Question

Hi ya

been doing an exercise and can'y get my head round this question.

A stone slides in a straight line across a frozen pond. Given that the intial speed of the stone is 5 ms and that it slides 20m before coming to rest. Calculate the coefficient of friction between the stone and surface of the frozen pond.

I worked out the the acceleration to be -0.625 but am not sure where to go next. Probably something really obvious.

Thanks

Matt

Santmaan2002 has got the right answer I just don't fully undertsand the working.

Answer 1

Initial velocity, u = 5 m/s
Distance, s = 20m
Final velocity, v = 0 m/s

Frictional force, f = (coeff of friction)(mg)
So acceleration, a = -(coeff of friction)(g) = -10(coeff of friction)

v^2 = u^2 + 2as
0 = 25 - 20(coeff of friction)(20)
400(coeff of friction) = 25
(coeff of friction) = 25/400
(coeff of friction) = 1/16 or 0.0625

Answer 2

hi Matt, i think of you're able to discover it a lot much less complicated to take an extremely diverse physique of strategies. start up with this equation: v² =vo² +2ax = vo² +2µkgx (the placement µ sub ok is the coefficient of kinetic friction. The letter "ok" is a subscript in all formula that stick to. Sorry, i don't understand a by using do subscripts.) µk = vo²/2gx =5/(29*8*20) =0.01275…. that's advisable to already comprehend this, yet merely in case you’re curious, the expression for “a” comes from this: fk =µkN and N = mg So, -µk = -µkmg = ma Then a =µkg want this facilitates. FE

Answer 3

Hi Matt,
I think you might find it easier to take a slightly different approach.
Start with this equation:
v² =vo² +2ax
= vo² +2µkgx (Where µ sub k is the coefficient of kinetic friction. The letter "k" is a subscript in all formulas that follow. Sorry, I don't know how to do subscripts.)
µk = vo²/2gx
=5/(29*8*20)
=0.01275….

You may already know this, but just in case you’re curious, the expression for “a” comes from this:
fk =µkN and N = mg
So,
-µk = -µkmg = ma
Then
a =µkg

Hope this helps.
FE

Answer 4

Use F = ma :) This gives you F in terms of m. But you know R in terms of m (i.e. R = mg, since the stone doesn't accelerate up or down). Using mu = F/R, the unknown mass term should cancel.

Edit: at the answer below, I don't think Work Done = Force * Distance has been covered yet. At the answer below that, the textbook uses 9.8 for g. This is Heinemann M1 (Edexcel) right? :)

Answer 5

Initial kinetic energy = 1/2 m*25
Friction force = u*mg and its work= u*mg *20
so
1/2m*25 = u*m*g*20 so u= 12.5/20 g =0.0638