1. A 2.15 kg block of wood is on a frictionless
inclined plane that makes an angle of 35.0^{o} with the
horizontal. a) Find F_{||}, and
F_{perp} b) If the block is
released on the plane, what will be its acceleration down the plane? c) What force in what direction would
prevent it from accelerating down the plane?

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a)

OK, basically,

F_{||} =
(2.15)*9.8*sin(35) = 12.085N = __12.1
N__

F_{perp} =
(2.15)*9.8*cos(35) = 17.260N = ** 17.3
N** but since
there is no friction the F

b)

Since there is no friction,
and no other things like jet engines, strings, or little men pushing or pulling
on the block, the only force along the plane is the F_{||} of 12.085N
down the plane (-). So our expression
of Newton’s second law becomes:

F = ma,

<-12.085 N> = (2.15
kg)a, (the parallel force is negative because down the plane is negative)

so a = -5.621 m/s/s = ** -5.62 m/s/s (down the plane)**.

c)

Since there is no friction
to help the block stay on the plane, you would need to cancel out the 12.085N
of weight parallel to the plane, and exert exactly that force up the
plane. Having so done, you would
prevent the block from accelerating up or down the plane, but it could still
move at a constant velocity in any direction.
__+12.1 N up the plane__

2. A 5.00 kg block of wood is on a frictionless 13 m
long inclined plane that makes an angle of 47^{o} with the
horizontal. a) Find F_{||}, and
F_{perp} b) If the block is
released on the plane, what will be its acceleration down the plane? c) What time will it take to slide down the
plane if it is released at the top from rest?
d) What force in what direction will make the block accelerate up the
plane at 4.50 m/s/s? e) What force in
what direction will make the block accelerate down the plane at 4.50 m/s/s?

Table of contents

a)

Again,

F_{||} =
(5.00)*9.8*sin(47) = 35.836N = __36
N__

F_{perp} =
(5.00)*9.8*cos(47) = 33.418N = ** 33
N** again, not
terribly useful.

b)

Since there is no friction Newton’s second law is simple. The only force acting along the plane is the parallel component of gravity down (-) the plane. Making down the plane negative:

<-35.836 N> = (5.00
kg)a, so a = -7.167 m/s/s = ** -7.2 m/s/s (down the
plane)**.

c)

s = -13, u = 0, a = -7.167
m/s/s, use s = ut + 1/2 at^{2}. t = 1.905 s = __1.9 s__

d)

So you have some unknown
force up the plane (+), and the parallel component of gravity 35.836N down (-)
the plane, and the acceleration is up (+) the plane, so your expression of
Newton’s Second law becomes

<+F – 35.836> = (5.00
kg)(+4.50 m/s/s) making up the plane positive.
F = 58.336N = __+58 N__

e)

Now we have the parallel force of 35.836 down the plane (-), and a force (F) that could be up or down the plane such that the box accelerates at 4.5 m/s/s down (-) the plane. (I know it needs a force up the plane, as the box, unfettered, will accelerate at 7.167 m/s/s down the plane) Newton’s second law gives us

<F – 35.836 N> = (5.00
kg)(-4.50 m/s/s) where the acceleration is negative because we are accelerating
down the plane, and up the plane is positive.
F = 13.336 N = ** +13 N**. Since it is positive, it is up the plane.

3. A block of Spam is on a frictionless inclined
plane that makes an angle of 35^{o} with the horizontal. a) If the Spam is released on the plane,
what will be its acceleration down the plane?
If the Spam is given a velocity of 5.0 m/s up the plane, b) how far up
the plane will the block slide before coming back down, and c) what time will
it take for the block to come back down to where it was released?

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a)

OK, we don’t know the mass
of the Spam, but we do know that it is tasty, and also some kind of meat. Don’t let this bother you. (Not knowing the
mass) The mass will cancel, or the
problem is unsolvable, except in terms of m.
Leave m in, and solve it that same way.

F_{||} =
m*9.8*sin(35) = 5.621m

F_{perp}
=m*9.8*cos(35) = 8.028m again, not terribly useful as the plane is
frictionless.

Since there is no friction the parallel force is the only force acting along the plane (down -) , so Newton’s second law is very simple to express:

<-5.621m> = ma, and m
cancels from all terms, giving a = -5.621m/s/s = ** -5.6 m/s/s (down the plane)** Pretty easy, huh?

b)

Solving linear kinematics,
the initial velocity is up the plane (+) and the acceleration is down (-)

s = ?, u = +5.00 m/s, a =
-5.621 m/s/s, v = 0 (when it reaches the top). t = ?

Use v = u + at, t = .890 s,

s = (u+v)/2*t, s = 2.224 m =
__+2.2 m__

c)

Since there is no friction,
the acceleration is the same for the block as it moves up or down the plane,
and so we can simply double the amount of time it took to go up the plane (it’s
like throwing a ball up in the air) so the time to return is t = 2*.890 = 1.78
s = ** 1.8 s**
(When there is friction, the block decelerates on the way up much more quickly
as it is opposed by the parallel component of gravity and friction, and
accelerates more slowly down the plane.
See problems 5-8)

4. A thing is on a frictionless 3.2 m long inclined
plane that makes an angle of 47^{o} with the horizontal. a) If the thing is released on the plane,
what will be its acceleration down the plane?
b) What time will it take to slide down the plane if it is released at
the top from rest?

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a)

OK, again, we don’t know the
mass of the thing, so let’s leave m in, and solve it that same way.

F_{||} =
m*9.8*sin(47) = 7.167m

F_{perp}
=m*9.8*cos(35) = 6.684m again, not terribly useful as the plane is
frictionless.

Since there is no friction,
the only force is the parallel force down (-) the plane. Our expression of
Newton’s law is simply:

<-7.167m> = ma, and m
cancels from all terms, giving a = -7.167 m/s/s = __-7.2 m/s/s (down the plane)__

b)

s = -3.2, u = 0, a = -7.167 m/s/s, v = ?, t = ?

Use s = ut + 1/2at^{2},
t = .9450 s = __.94 s__

5. A 5.00 kg block of wood is on an inclined plane that
makes an angle of 32.0^{o} with the horizontal. There is a static coefficient of friction of
.670 and a kinetic of .340 between the block and the plane. a) Find F_{||}, F_{perp},
maximum F_{fstatic}, and F_{fkinetic} b) If the block initially
at rest on the plane, will it start to slide down the plane all by itself? What
is the force of static friction that exists between the plane and the block
when the block is at rest on the plane?
c) What additional force is needed to make the block start to slide down
the plane? d) Once it is started down
the plane, what is its acceleration down the plane? e) What force is needed to make the block start sliding up the
plane if it is initially at rest? f)
What force would make the block slide down the plane with an acceleration of
8.30 m/s/s? g) What force would make
the block slide up the plane with an acceleration of 2.55 m/s/s?

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a)

F_{||} =
(5.00)*9.8*sin(32) = 25.966 N = __26.0
N__

F_{perp} =
(5.00)*9.8*cos(32) = 41.554 N = ** 41.6
N**– This force
becomes the normal force, or the force holding the block against the plane, and
is used to calculate the friction force F

F_{fstatic(max)} =
μ_{s}F_{perp} =
(.67)(41.554 N) = 27.841N = ** 27.8
N **(This is a
maximum value)

F_{fkinetic} =
μ_{k}F** _{perp} =**(.34)(41.554 N) = 14.128 N =

b)

If the block is sitting
still on the plane, then there could be a static force of friction as high as
27.841 N holding it in place. The F_{||}
of 25.966 N is trying to make it slide down the plane, which is not enough, so
the block sits on the plane with a force of 25.966 N down the plane,
counteracted by a static frictional force of 25.966 N = ** +26.0 N up** the plane. (Static friction is a “up to this amount”
force. In this case, the static
friction will go up to 27.841N, but no higher.)

c)

You would need to add a
force down the plane equal to the difference between the parallel component of
gravity (25.966) and the maximum static frictional force. (27.841)
So you would need 27.841-25.966 = -1.875 N = ** -1.9 N more (down the plane)**.

d)

Once it is moving down the
plane, the static friction force is no longer relevant, and so we have the F_{fkinetic}
of 14.128N up (+) the plane, (friction is in the opposite direction of motion
when you are dragging something, so since we are going down the plane, it would
be up the plane) and the F_{||} of 25.966N down (-) the plane. Our expression of F = ma looks like

<14.128 N– 25.966 N> =
(5.00 kg)a where up the plane is positive, and a = -2.368 m/s/s = __-2.37 m/s/s (down the plane)__

e)

You would need to overcome
the static friction, and the parallel force, so you would need 27.841 + 25.966
= 53.8075 N = __+55.8 N up the
plane.__

f)

Making up the plane positive, the acceleration is down the plane and therefore negative, and we have the kinetic friction acting up (+) the plane opposite the velocity, (assuming the block is moving – which is a good assumption since we know its acceleration) and the parallel force down (-) the plane and some applied force (F), so F = ma looks like:

<14.128 N– 25.966 N +
F> = (5.00 kg)(-8.30 m/s/s), (Where up the plane is positive) and F = -29.66
N, or __-29.7 N (down the plane)__** ,** which makes sense because
the block is wont to accelerate at only 2.368 m/s/s without any help.

g)

This is very similar to the previous problem, except
now the kinetic friction force of 14.128 N is __down__ (-) the plane, as the
block is sliding __up__ the plane, and there is the parallel force, as
always, down (-) the plane. F = ma
looks like this where F is the unknown applied force:

<F – 25.966 N – 14.128
N> = (5.00 kg)(+2.55 m/s/s) (Again making up the plane positive)

so F = 52.84 N = __+52.8 N up the plane.__

6. A 12.0 kg block of wood is on an inclined plane
that makes an angle of 55^{o} with the horizontal. There is a static coefficient of friction of
.85 and a kinetic of .65 between the block and the plane. a) Find F_{||}, F_{perp},
maximum F_{fstatic}, and F_{fkinetic} b) If the block initially
at rest on the plane, will it start to slide down the plane all by itself? c) Once it is started down the plane, what
is its acceleration down the plane? d)
What force is needed to make the block start sliding up the plane if it is
initially at rest? e) What force would
make the block slide down the plane with an acceleration of 1.50 m/s/s? f) What force would make the block slide up
the plane with an acceleration of 5.4 m/s/s?

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a)

F_{||} =
(12.0)*9.8*sin(55) = 96.3323N = __96
N__

F_{perp} =
(12.0)*9.8*cos(55) = 67.4526N = ** 67
N**– This force
becomes the normal force, or the force holding the block against the plane, and
is used to calculate the friction force F

F_{fstatic(max)} =
μ_{s}F_{perp} =
.85*67.4526N = 57.3347N = ** 57 N** (This is a maximum value)

F_{fkinetic} =
μ_{k}F** _{perp} =**.65*67.4526N N = 43.8442N =

b)

Since the F_{||}
> F_{fstatic(max)} in this
case it will start to slide all by itself.

c)

Once it is moving down the
plane, the static friction force is no longer relevant, and so we have the F_{fkinetic}
of 43.8442N up (+) the plane, and the F_{||} of 96.3323N down (-) the
plane. Our expression of F = ma looks
like <43.8442N – 96.3323N> = (12.0 kg)a where up the plane is positive,
and a = -4.374 m/s/s = __-4.4
m/s/s (down the plane)__

d)

You would need to overcome
the static friction, and the parallel force, so you would need 57.3347N +
96.3323N = 153.667 N = __+150 N
up the plane__.

e)

Making up the plane positive, as we slide down the plane, we have the kinetic friction up (+) the plane, and the parallel force (always down), so F = ma looks like:

<43.8442 N – 96.3323 N + F> = (12.0 kg)(-1.50
m/s/s), (Where up the plane is positive – the acceleration is negative as it is
down the plane) and F = 34.488 N, or ** +34 N up the plane**,
which makes sense because the block will accelerate faster than 1.50 m/s/s
(4.374 m/s/s) down the plane unassisted.

f)

If we are sliding up the
plane, the kinetic friction now acts __down__ (-) the plane. Our other players along the plane are the
applied force (F) and the parallel force down (-) the plane. So F = ma looks like:

<-43.8442N – 96.3323N +
F> = (12.0 kg)(+5.40
m/s/s), (Where up the plane is positive, so the acceleration is positive this
time) and F = 204.977 N, or ** +2.0
x 10^{2} N up the plane**.

7 It takes a force of 275 N to slide a 26.0 kg box
up a 50.0^{o} inclined plane at a constant velocity. a) What is the
force of friction opposing the motion?
b) What is the coefficient of friction between the box and the
plane? c) If the box were to slide
freely down the plane, what would be its acceleration? d) What force would make the box slide up
the plane with an acceleration of 6.2 m/s/s?

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a)

F_{||} =
(26.0)*9.80*sin(50) = 195.1881 N

F_{perp} =
(26.0)*9.80*cos(50) = 163.7823 N

If it is sliding up the
plane with no acceleration, then the 275 N force up (+) the plane must exactly
equal the parallel component of the weight down the plane (-) plus the friction
force, so Newton’s second law looks like:

<+275 N – 195.1881 N +
F> = (26.0 kg)(0 m/s/s) (up the plane is +, and F must be the force of
friction – the acceleration is zero as it is moving at a constant velocity, and
not accelerating)

F = -79.8119 N = __-79.8 N (down the plane)__

b)

Since this is the friction
force, we can use the formula for kinetic friction to find the coefficient:

F_{fkinetic} =
μ_{k}F_{perp}

79.8119 N = μ_{k}(163.7823
N)

μ_{k} = .4873 =
__.487__

c)

Were the box freely sliding,
the parallel force would be acting down (-) the plane, and the friction force
would oppose it and act up the plane (+) , so Newton’s second law looks like:

<-195.1881 N + 79.8119
N> = (26.0 kg)a

a = -4.4375 m/s/s =** -4.44 m/s/s (down the plane)**

d) Sliding up the plane, the
parallel force would still be down (-) the plane, friction would be opposite
the velocity and therefore down (-) the plane, and the applied force (F) would
be up the plane (+). The acceleration
would be positive, as it is up the plane.:

<F - 195.1881 N - 79.8119
N> = (26.0 kg)(+6.2 m/s/s)

F = 436.2 N = __+436 N up the plane __

8. A piece of candy is on an inclined plane that
makes an angle of 23^{o} with the horizontal. There is a static coefficient of friction of .72 and a kinetic of
.30 between the block and the plane. a)
If the block is initially at rest on the plane, will it start to slide down the
plane all by itself? b) Once it is
started down the plane, what is its acceleration down the plane? If it is given a shove up the plane at 4.15
m/s, c) how far up the plane does it go, d) and what time does it take to get
back down to where it was shoved (assuming it does not stick at the top), and
e) what is its velocity when it returns?

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a)

You know the drill:

F_{||} =
m9.8*sin(23) = 3.8292m

F_{perp} =
m9.8*cos(23) = 9.0209m– This force becomes the normal force

F_{fstatic(max)} =
μ_{s}F_{perp} =
.72(9.0209m) = 6.4951m (This is a maximum value)

F_{fkinetic} =
μ_{k}F** _{perp} =**.30(9.0209m) N = 2.7063m

So, the piece of candy will
not start on its own as the parallel force of (3.8292m) is smaller than the
maximum static force of friction (6.4951m). __No__

b)

Sliding down the plane, friction is up the plane (+), and parallel down (-), so F = ma looks like:

<-3.8292m + +2.7063m>
= ma so a = -1.1229 m/s/s or __-1.1
m/s/s (down the plane)__

c)

As it slides up the plane, friction acts down the plane (-) opposite the motion, and the parallel (wie immer) is down (-) the plane:

<-3.8292m + -2.7063m> = ma, a = -6.5354 m/s/s

s = ?, u = +4.15 m/s, v = 0
(at top), a = -6.5354 m/s/s, t = ?

use v^{2} = u^{2}
+ 2as, s = 1.3176 m = __+1.3 m
up the plane__

d)

s = 1.3176 m, u = +4.15 m/s,
v = 0 (at top), a = -6.5354 m/s/s, t = ?

use v = u + at, t = .6350 s
(for the trip up the plane)

Now we have to solve for the
trip back down the plane. We have already figured out the acceleration down the
plane:

s = -1.3176 m (i.e. we go back down the same distance we went up), u = 0, v = ?, a = -1.1229 m/s/s, t = ?

use s = ut + 1/2at^{2},
t = 1.5319 s, so the round trip is .6350 s (up) + 1.5319 s (down) = 2.1669 s = __2.2 s__

e)

The final velocity at the
bottom:

s = -1.3176 m, u = 0, v = ?, a = -1.1229 m/s/s, t = 1.5319 s

Use v = u + at, v = -1.7202
= ** -1.7 m/s (down the plane) **– it does not return with
the same velocity as it has lost energy to friction.

9. A crate is sliding at 3.20 m/s down a 27.6^{o}
plane and is stopped by friction in a distance of 4.22 m. a) What is the coefficient of friction
between the crate and the plane?

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First we have a cute linear kinematics problem to solve. (Where else could we start??)

s = -4.22 m (down the
plane), u = -3.20 m/s, v = 0 (stops), a = ???, t = don’t care

use v^{2} = u^{2}
+ 2as, a = +1.2133 m/s/s(up the plane)

Second we need to express
the normal force, parallel force and kinetic friction in terms of m:

F_{||} =
m9.8*sin(27.6) = 4.5403m

F_{perp} =
m9.8*cos(27.6) = 8.6848m– This force becomes the normal force

F_{fkinetic} =
μ_{k}F** _{perp} =**μ

Third – set up Newton’s
second law. As it slides down the
plane, there is the parallel force down the plane (-) and the friction force up
the plane (+) and the acceleration is up the plane (+) as it is stopping:

<-4.5403m + μ_{k}8.6848m>
= m(+1.2133 m/s/s) - m divides out from both sides leaving:

<-4.5403 + μ_{k}8.6848>
= +1.2133 m/s/s

μ_{k} = (4.5403
+ 1.2133)/8.6848 __= .662__