 # Solutions to Sample Problems 1.

```Solutions to Sample Problems
1.
What is the force of friction between a block of ice that weighs 930 N and the ground if Mu =
.12? (111.6 N)
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W = 930 N
Mu = .12
Once again, our formula for Mu is:
F = Mu (FN)
It's pretty easy to plug-in the stuff we know:
F = (.12)(930 N)
We then do the math to get 111.6 N.
2.
What is the coefficient of static friction if it takes 34 N of force to move a box that weighs 67 N?
(.51)
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F = 3.0 s
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W = 67 N
Once again, our formula for Mu is:
F = Mu (FN)
We know that FN is equal to W so we can plug what we know into the formula and
get:
34 N = MuS (67 N)
Divide the 34 by 67 and you end up with .507462 which happens to round to the
.51.
3.
A box takes 350 N to start moving when the coefficient of static friction is .35. What is the
weight of the box? (1000 N)
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F = 350
Mu = .35
Hey, since FN = W, surprisingly enough we use the formula:
F = Mu (FN)
which we plug-in the numbers and get
350 = .35 (FN)
Divide 350 by .35 to get 1000 N.
4.
A car has a mass of 1020 Kg and has a coefficient of friction between the ground and its tires of
.85. What force of friction can it exert on the ground? What is the maximum acceleration of this
car? In what minimum distance could it stop from 27 m/s? (8500 N, 8.3 m/s/s, 43.8 m)
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m = 1020 Kg
Mu = .85
Use the formula
W = mg
with g = 9.8 m/s/s as the force of gravity. We now know the weight to be 9996
N. We can
then use the formula
F = Mu (FN) with FN equal to the weight.
Plugged in it looks like:
F = (.85)(9996 N) and equals 8496.6 N but with significant digits it ends up
being 8500
N. In order to find the acceleration we need a formula that uses the Force we
found and
acceleration. Obviously
F = ma
seems best. We plug-in the numbers and get:
8496.6 N = (1020 Kg) a
Solve for a and get 8.33 m/s/s as the acceleration
But we're not done yet! We need to solve for the distance. Since this deals
with good old
velocity, acceleration and distance we need a formula that uses only what we
know:
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Vo = 37 m/s
V = 0 (It comes to a complete stop)
a = 8.33 m/s/s
Now we can use:
x = V2 - VO 2
----------------2a
x = 43.7575 or with sig figs, 43.8 m
5.
Clarice moves a 800 gram set of weights by applying a force of 1.2 N. What is the coefficient of
friction? (.15)
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m = 800 grams or .8 Kg
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F = 1.2 N
g = 9.8 m/s/s
We first need to get the FN or Weight; we must convert the mass to kilograms
and then get FN = 7.84 N for the next step.
As usual when we look at the formulas we see that:
F = Mu (FN)
seems to be the only one that will uses Mu
We plug-in the numbers to get:
1.2 = Mu (7.84 N)
Divide 1.2 by 7.84 to get .15306 or .15 for the coefficient of friction.
6.
A car has a coefficient of friction between the ground and its tires of .85. What is the mass of the
car if it takes 9620 N of force to make it slide along the ground? (1155 Kg)
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Mu = .85
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F = 9620 N
We have to find the weight first so we use:
F = Mu (FN) and that equals
9620 N = (.85)(FN)
9620 divided by .85 = 11317.647 N
But we need this in mass so we use:
F = mg
11317.647 = (m)(9.8 m/s/s)
so the mass equals 1154.8619 or 1155 kg
7.
A 5.0 Kg block has a coefficient of friction of .15 on a flat surface. What is its acceleration if you
exert a force of 15 N sideways on it when it is at rest? (Find the friction force first) (1.47 m/s/s)
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m = 5.0 kg
F = 15 N
Mu = .15
So we can use the formula:
FN = mg or 15 N = (5.0 kg)(9.8 m/s/s)
So the FN = 49 N
Then we plug that into:
F = Mu (FN) that looks like
F = (.15)(49 N)
And get the actual force to be 7.35 N
And then we can plug that back into the formula:
F = ma or 7.35 N = (5.0)a
Divide 7.35 by 5 and get 1.47 m/s/s
8.
A 10 Kg block is at rest on a level surface. It accelerates from rest to 51.2 m/s in 8 seconds when
you exert a force of 100 N on it sideways. What is the acceleration of the block? What is the
force of friction between the surface and the block, and what is the coefficient of friction? (6.4
m/s/s 36 N, .56)
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m = 10 Kg
Vo = 0 m/s
V = 51.2 m/s
t = 8 seconds
F = 100 N
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Find the acceleration using the our freindly neighborhood aceleration formula
once
again:
V = Vo + at.
Plug in a couple of numbers and get:
51.2 m/s = 0 m/s + (a * 8 s)
Solve for a, and we get 6.4 m/s/s
The force of friction = Force applied ñ The normal force
FA = 100 N and FN = 64 N so the FF = 36 N
Now we plug that back into the formula:
FF = Mu (FN) we use our stuff
36 N = Mu (64 N)
Divide 36 by 64 and get .5625 as the coefficient of friction.
9.
A 120 Kg log sled accelerates at 1.4 m/s/s when a horse pulls on it. What force must the horse
exert if the coefficient of friction between the ground and the sled is .28? (497 N)
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m = 120 Kg
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Mu = .28
a = 1.4 m/s/s
First we need to find FN and we use FN = mg
F = (120 Kg)(9.8 m/s/s) and F equals 1176 N
We then multiply that by the coefficient of friction (.28)to get the force of
friction
which is 329.28 N. We then use the formula:
F = FA - FF, with the values plugged in we get:
168 = FA - 329.28
solve this little problem and get 497.28 or 497 N.
10.
(E.C. - Super Stud Problem) You exert a force of 24 N sideways on an object and it accelerates
from 0 - 12 m/s over a distance of 5.2 m. You know that the coefficient of friction between the
object and the ground is .58, so what is its mass? (1.23 Kg)
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VO = 0 m/s
V = 12 m/s
Mu = .58
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x = 5.2 m
FA = 24 N
Letís use x = V2 - VO2
------------2a
we plug in what we know and get:
5.2 m = 122 -
2
/ 2a
solve for acceleration and get a = 13.846 m/s/s
we want to find F = ma but weíll need to use:
F = FA - FF but we donít know the FF
So FF = Mu * FN we plug stuff in and get:
Since F = ma and FF = Mu * m * g we can put it into
F = FA - FF so it looks like:
ma = FA - (Mu * m * g) we can then put in what we know
m (13.846) = 24 N ñ (.58 * m * 9.8m/s/s) Wait I donít Know!
13.846 = 24/m ñ 5.684 add 5.684m to both sides and NO That doesnít Work!
6.684m * (13.846) = 24 OHHH MAN!
Ohhh dang it this just ends up being 1.23 Kg!!!!!
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