Final Exam Review Semester

1
Final Exam ReviewSemester 1
I hope this will help you to prepare for your
Final exam in Honors Physics. This presentation
will present a series of problems. Work through
the problems on paper, than click to see the
solutions.
Click the mouse to go to problem 1
2
PROBLEM 1
One light year is the distance light travels in
one year. Calculate the distance to the nearest
star which is 4.5 light years away ? The speed
of light is 3 x 108 m/s.
Start by identifying the given information.(The
Data)
Click to Advance
3
PROBLEM 1
One light year is the distance light travels in
one year. Calculate the distance to the nearest
star which is 4.5 light years away ? The speed
of light is 3 x 108 m/s.
vL 3 x 108 m/s t 4.5 Light Years
Next determine the appropriate formula to Solve
the problem.
Click to Advance
4
PROBLEM 1
One light year is the distance light travels in
one year. Calculate the distance to the nearest
star which is 4.5 light years away ? The speed
of light is 3 x 108 m/s.
vL 3 x 108 m/s t 4.5 Light Years
Since light travel at a constant speed, this is
a uniform motion problem. Thus v d/t or d vt
Also the speed of light is measured in m/s , so
a light year must be expressed in light seconds
in order to do the problem. Do this calculation
before your move on !
Click to Advance
5
PROBLEM 1
One light year is the distance light travels in
one year. Calculate the distance to the nearest
star which is 4.5 light years away ? The speed
of light is 3 x 108 m/s.
vL 3 x 108 m/s t 4.5 Light Years
4.5 Lt.yr x 365.25 da/yr x 24 hr/da x 3600 s/hr
1.42 x 108 Lt.s
Now convert Light Seconds to distance in Meters.
Click to Advance
6
PROBLEM 1
One light year is the distance light travels in
one year. Calculate the distance to the nearest
star which is 4.5 light years away ? The speed
of light is 3 x 108 m/s.
vL 3 x 108 m/s t 4.5 Light Years
1.42 x 108 Lt.s x 3 x 108 m/ Lt.s
4.26 x 1016 meters
Click to Advance To Next Problem
7
PROBLEM 2
A High Jumper is going for the new worlds record
of 2.52 m. At what speed would he have to leave
the ground in order to rise to this height and
clear the bar ?
Start by identifying the given information.(The
Data)
Click to Advance
8
PROBLEM 2
A High Jumper is going for the new worlds record
of 2.52 m. At what speed would he have to leave
the ground in order to rise to this height and
clear the bar ?
d 2.52 m v2 0 m/s g -9.81 m/s2 v1 ?
Next determine the appropriate formula to Solve
the problem.
Click to Advance
9
PROBLEM 2
A High Jumper is going for the new worlds record
of 2.52 m. At what speed would he have to leave
the ground in order to rise to this height and
clear the bar ?
d 2.52 m v2 0 m/s g -9.81 m/s2 v1 ?
Motion up and down is uniform accelerated motion
thus one of the three formulas for UAM applies.
Based on the given data you can use v22 v12
2gd.
Rearrange the formula to solve for v1.
Click to Advance
10
PROBLEM 2
A High Jumper is going for the new worlds record
of 2.52 m. At what speed would he have to leave
the ground in order to rise to this height and
clear the bar ?
d 2.52 m v2 0 m/s g -9.81 m/s2 v1 ?
Subtracting 2gd from both sides yields V12 v22
- 2gd, now v22 will cancel out because it equals
0, thus v1 v-2gd
Plug in the data and solve for v1.
Click to Advance
11
PROBLEM 2
A High Jumper is going for the new worlds record
of 2.52 m. At what speed would he have to leave
the ground in order to rise to this height and
clear the bar ?
d 2.52 m v2 0 m/s g -9.81 m/s2 v1 ?
v1 v-2gd v-(2)(-9.81 m/s2)(2.25m) v1
6.64 m/s
Click to Advance To The Next Problem
12
PROBLEM 3
An Archer shoot an arrow from a cliff with a
velocity of 20 m/s,it strikes the ground in 10 s.
How high is the cliff, and how far from the base
of the cliff does the arrow land ?
Start by identifying the given information.(The
Data)
Click to Advance
13
PROBLEM 3
An Archer shoot an arrow from a cliff with a
velocity of 20 m/s, it strikes the ground in 10
s. How high is the cliff, and how far from the
base of the cliff does the arrow land ?
t 10 s v1 20 m/s R ? g 9.81 m/s2 h ?
Next determine the appropriate formula to Solve
the problem.
Click to Advance
14
PROBLEM 3
An Archer shoot an arrow from a cliff with a
velocity of 20 m/s, it strikes the ground in 10
s. How high is the cliff, and how far from the
base of the cliff does the arrow land ?
t 10 s v1 20 m/s R ? g 9.81 m/s2 h
? vy1 0 m/s
This is a 1/2 parabola problem since no
trajectory angle is given. The value of g is
positive moving downward. Use the formula d
vy1t 1/2gt2, to solve for d (h).
Plug in the data and solve for d.
Click to Advance
15
PROBLEM 3
An Archer shoot an arrow from a cliff with a
velocity of 20 m/s, it strikes the ground in 10
s. How high is the cliff, and how far from the
base of the cliff does the arrow land ?
t 10 s v1 20 m/s R ? g 9.81 m/s2 h
? vy1 0 m/s
Since vy1 0 m/s it cancels out of the formula
thus h 1/2gt2 h 1/2 (9.81 m/s2)(10s)2
490.5 m high.
Next solve for R.
Click to Advance
16
PROBLEM 3
An Archer shoot an arrow from a cliff with a
velocity of 20 m/s, it strikes the ground in 10
s. How high is the cliff, and how far from the
base of the cliff does the arrow land ?
t 10 s v1 20 m/s R ? g 9.81 m/s2 h
? vy1 0 m/s
Since v1 is only in the x direction, v1 vx in
the formula for Range, R vxT and since it is a
1/2 parabola problem t T. Thus R (20
m/s)(10s) 200 m.
Click to Advance
17
PROBLEM 4
A baseball is thrown downward at 2 m/s from a
deck on the Eiffel tower, 175 m above the
ground. Disregarding air resistance a) How
long does it take for the ball to hit the
ground, and b) How fast is the ball going as it
hits the ground ?
Start by identifying the given information.(The
Data)
Click to Advance
18
PROBLEM 4
A baseball is thrown downward at 2 m/s from a
deck on the Eiffel tower, 175 m above the
ground. Disregarding air resistance a) How
long does it take for the ball to hit the
ground, and b) How fast is the ball going as it
hits the ground ?
vy1 2 m/s d 175 m vy2 ? g 9.81 m/s2
t ?
Next determine the appropriate formula to Solve
the problem.
Click to Advance
19
PROBLEM 4
A baseball is thrown downward at 2 m/s from a
deck on the Eiffel tower, 175 m above the
ground. Disregarding air resistance a) How
long does it take for the ball to hit the
ground, and b) How fast is the ball going as it
hits the ground ?
vy1 2 m/s d 175 m vy2 ? g 9.81 m/s2
t ?
Time can be calculated from the formula d vy1t
1/2gt2. Since vy1 is not 0, the entire formula
must be used. Rearrange to solve for t.
Click to Advance
20
PROBLEM 4
A baseball is thrown downward at 2 m/s from a
deck on the Eiffel tower, 175 m above the
ground. Disregarding air resistance a) How
long does it take for the ball to hit the
ground, and b) How fast is the ball going as it
hits the ground ?
vy1 2 m/s d 175 m vy2 ? g 9.81 m/s2
t ?
This yields 1/2gt2 vy1t - d 0. A
Quadratic Equation ! Using the program in your
calculators to solve the quadratic equation.
Click to Advance
21
PROBLEM 4
A baseball is thrown downward at 2 m/s from a
deck on the Eiffel tower, 175 m above the
ground. Disregarding air resistance a) How
long does it take for the ball to hit the
ground, and b) How fast is the ball going as it
hits the ground ?
vy1 2 m/s d 175 m vy2 ? g 9.81 m/s2
t ?
1/2(9.81 m/s2 )t2 (2 m/s )t - 175m 0 (4.90
m/s2 )t2 (2 m/s )t - 175m 0 This results in
two values for t 5.78 s -6.18 s
Since time can never be negative, 5.78 s must be
correct. Next Calculate v2.
Click to Advance
22
PROBLEM 4
A baseball is thrown downward at 2 m/s from a
deck on the Eiffel tower, 175 m above the
ground. Disregarding air resistance a) How
long does it take for the ball to hit the
ground, and b) How fast is the ball going as it
hits the ground ?
vy1 2 m/s d 175 m vy2 ? g 9.81 m/s2
t 5.78 s
Final velocity can be calculated from a
v2 - v1 Rearrange and solve. t
Click to Advance
23
PROBLEM 4
A baseball is thrown downward at 2 m/s from a
deck on the Eiffel tower, 175 m above the
ground. Disregarding air resistance a) How
long does it take for the ball to hit the
ground, and b) How fast is the ball going as it
hits the ground ?
vy1 2 m/s d 175 m vy2 ? g 9.81 m/s2
t 5.78 s
The equation should read v2 v1 at Thus
v2 2 m/s (9.81 m/s2 )(5.78 s) 58.7 m/s
Click to Advance To Next Problem
24
PROBLEM 5
A box is being pulled across the floor by a
rope attached at a 40o angle. A force of 150 n.
is applied to the rope. Calculate the x and y
components of the 150 n. force.
Start by drawing a force diagram of the give
situation.
Click to Advance
25
PROBLEM 5
A box is being pulled across the floor by a
rope attached at a 40o angle. A force of 150 n.
is applied to the rope. Calculate the x and y
components of the 150 n. force.
Fa 150 n ? 40o
Fa
Fy
?
Fx
Using basic Trig. Calculate Fx and Fy.
Click to Advance
26
PROBLEM 5
A box is being pulled across the floor by a
rope attached at a 40o angle. A force of 150 n.
is applied to the rope. Calculate the x and y
components of the 150 n. force.
150 n.
Fy
?
Fx
Fx cos?F cos40o(150 n) 114.9 n. Fy.
sin?F sin40o(150 n) 96.42 n.
Click to Advance To The Next Problem
27
Vector Analysis
To review Vector analysis Download the Power
Point Presentation on Right Angle Systems and
Non Right Angle Systems from the web site and
work through the problems provided there.
Click to Advance To The Next Problem
28
PROBLEM 6
A football player, punts the ball with a
trajectory velocity of 12 m/s at an angle of 55o
above the horizontal. Calculate the hang time,
and the distance the ball travels down field.
Start by drawing a diagram of the give
situation, including all vectors.
Click to Advance
29
PROBLEM 6
A football player, punts the ball with a
trajectory velocity of 12 m/s at an angle of 55o
above the horizontal. Calculate the hang time,
and the distance the ball travels down field.
vT
vy
?
vx
Followed by identifying the given
information.(The Data)
Click to Advance
Click to Advance
30
PROBLEM 6
A football player, punts the ball with a
trajectory velocity of 12 m/s at an angle of 55o
above the horizontal. Calculate the hang time,
and the distance the ball travels down field.
vT
vy1
?
vx
vT 12 m/s ? 55o vy2 0 m/s g -9.81 m/s2
? g 9.81 m/s2 ? t ? R ?
Next Calculate the component vectors
Click to Advance
Click to Advance
31
PROBLEM 6
A football player, punts the ball with a
trajectory velocity of 12 m/s at an angle of 55o
above the horizontal. Calculate the hang time,
and the distance the ball travels down field.
vT 12 m/s ? 55o g -9.81 m/s2 ? g
9.81 m/s2 ? vy2 0 m/s t ?
R ?
vT
vy1
?
vx
vy1 sin?vT sin55o(12 m/s) 9.83 m/s vx
cos?vT cos55o(12 m/s) 6.88 m/s
Now focus on the Hang time
Click to Advance
Click to Advance
32
PROBLEM 6
A football player, punts the ball with a
trajectory velocity of 12 m/s at an angle of 55o
above the horizontal. Calculate the hang time,
and the distance the ball travels down field.
vT 12 m/s ? 55o g -9.81 m/s2 ?
g 9.81 m/s2 ? vy1 9.83 m/s vx 6.88
m/s vy2 0 m/s t ?
R ?
vT
vy1
?
vx
Since g? is (-), t? can be calculated from t
vy2 - vy1 - 9.83 m/s 1 s.
g - 9.81 m/s2
Doubling this value gives you the total Hang
Time T 2t 2 s.
Click to Advance
33
PROBLEM 6
A football player, punts the ball with a
trajectory velocity of 12 m/s at an angle of 55o
above the horizontal. Calculate the hang time,
and the distance the ball travels down field.
vT 12 m/s ? 55o g -9.81 m/s2 ?
g 9.81 m/s2 ? vy1 9.83 m/s vx 6.88
m/s vy2 0 m/s t 1 s. T 2 s
R ?
vT
vy1
?
vx
The Range can be calculated from R vx T
(6.88 m/s)(2 s) 13.76 m Thus, Hang Time
2 s. Range 13.76 m
Click to Advance To The Next Problem
34
PROBLEM 7
A box having a mass of 50 kg is dragged across
a horizontal floor at a constant speed by means
of a rope tied on the front of it. A force of
70 n is applied to the rope which makes an angle
of 30o with the horizontal. Calculate the
coefficient of friction.
Start by drawing a Force Diagram of the give
situation, including all Force Arrows. Identify
the given data.
Click to Advance
35
PROBLEM 7
A box having a mass of 50 kg is dragged across
a horizontal floor at a constant speed by means
of a rope tied on the front of it. A force of
70 n is applied to the rope which makes an angle
of 30o with the horizontal. Calculate the
coefficient of friction.

• m 50 Kg
• µ ?
• 30o
• Fa 70 n

Fa
Fy
Ff
?
Fx
Fn
Determine the equations necessary to solve the
problem.
Click to Advance
36
PROBLEM 7
A box having a mass of 50 kg is dragged across
a horizontal floor at a constant speed by means
of a rope tied on the front of it. A force of
70 n is applied to the rope which makes an angle
of 30o with the horizontal. Calculate the
coefficient of friction.

• m 50 Kg
• µ ?
• 30o
• Fa 70 n

Fa
Fy
Ff
?
Fx
Fn
Start at the end of the problem µ Ff/Fn.
Now Ff Fx (UM) and Fn mg - Fy , so
calculating the x and y components is the first
step.
Click to Advance
37
PROBLEM 7
A box having a mass of 50 kg is dragged across
a horizontal floor at a constant speed by means
of a rope tied on the front of it. A force of
70 n is applied to the rope which makes an angle
of 30o with the horizontal. Calculate the
coefficient of friction.

• m 50 Kg
• µ ?
• 30o
• Fa 70 n

Fa
Fy
Ff
?
Fx
Fn
Fx cos?Fa cos30o(70 n) 60.62 n. Fy
sin?Fa sin30o(70 n) 35.00 n. Also
mg (70 Kg)(9.81 m/s2) 686.7 n. Plug these
values into your formulas.
Click to Advance
38
PROBLEM 7
A box having a mass of 50 kg is dragged across
a horizontal floor at a constant speed by means
of a rope tied on the front of it. A force of
70 n is applied to the rope which makes an angle
of 30o with the horizontal. Calculate the
coefficient of friction.

• m 50 Kg mg 686.7 n
• µ ? Fx 60.62 n
• 30o Fy 35.00 n
• Fa 70 n

Fa
Fy
Ff
?
Fx
Fn
µ Ff Fx 60.62 n
60.62 n Fn mg - Fy
686.7 n - 35 n 651.7 n µ .093
Click to Advance To Next Problem
39
PROBLEM 8
Calculate the force required to accelerate a 200
Kg. box across the floor at 5 m/s2, if the force
is applied parallel to the ground, and the
coefficient of friction is .5. How far will the
box move if it starts from rest And is pulled for
3.5 seconds ?
Start by drawing a Force Diagram of the give
situation, including all Force Arrows. Identify
the given data.
Click to Advance
40
PROBLEM 8
Calculate the force required to accelerate a 200
Kg. box across the floor at 5 m/s2, if the force
is applied parallel to the ground, and the
coefficient of friction is .5. How far will the
box move if it starts from rest And is pulled for
3.5 seconds ?

• m 200 Kg
• µ .50
• a 5 m/s2
• Fa ?

Fnet
Fa
Ff
Fn
Determine the equations necessary to solve the
problem.
Click to Advance
41
PROBLEM 8
Calculate the force required to accelerate a 200
Kg. box across the floor at 5 m/s2, if the force
is applied parallel to the ground, and the
coefficient of friction is .5. How far will the
box move if it starts from rest And is pulled for
3.5 seconds ?
Fnet

• m 200 Kg
• µ .50
• a 5 m/s2
• Fa ?

Fa
Ff
Fn
Working from the end of the problem Fnet ma
(Newton II), also Fnet Fa - Ff, Therefore Fa
Fnet Ff, and Ff µFn, also Fn mg
(Flat surface). So Calculating mg and Fnet is
the first step.
Click to Advance
42
PROBLEM 8
Calculate the force required to accelerate a 200
Kg. box across the floor at 5 m/s2, if the force
is applied parallel to the ground, and the
coefficient of friction is .5. How far will the
box move if it starts from rest And is pulled for
3.5 seconds ?
Fnet

• m 200 Kg
• µ .50
• a 5 m/s2
• Fa ?

Fa
Ff
Fn
Fnet ma (200 Kg)(5 m/s2) 1000 n mg (200
Kg)(9.81 m/s2) 1962 n Plug these values into
your formulas.
Click to Advance
43
PROBLEM 8
Calculate the force required to accelerate a 200
Kg. box across the floor at 5 m/s2, if the force
is applied parallel to the ground, and the
coefficient of friction is .5. How far will the
box move if it starts from rest And is pulled for
3.5 seconds ?
Fnet

• m 200 Kg
• µ .50
• a 5 m/s2
• Fa ?
• Fnet 1000 n
• mg 1962 n

Fa
Ff
Fn
Fa Fnet Ff 1000 n - µ mg Fa 1000 n µFn
Fa 1000 n (.5)(1962 n) Fa 1000 n 981 n
1981 n
Now focus on the distance traveled.
Click to Advance
44
PROBLEM 8
Calculate the force required to accelerate a 200
Kg. box across the floor at 5 m/s2, if the force
is applied parallel to the ground, and the
coefficient of friction is .5. How far will the
box move if it starts from rest And is pulled for
3.5 seconds ?
Fnet

• m 200 Kg
• µ .50
• a 5 m/s2
• Fa ?
• Fnet 1000 n
• mg 1962 n

Fa
Ff
Fn
Since the time is given as 3.5 seconds, d can be
calculated from d v1t 1/2at2 Since v1 0 it
cancels out leaving d 1/2at2 1/2 (5
m/s2)(3.5s)2 30.63 m
Click to Advance To The Next Problem
45
PROBLEM 9
A Helicopter weighs 2.0 x 103 N. Its
rotating blades exert 2.5 x 103 N. of force as
it lifts-off the ground. (a) What is the
helicopters acceleration during lift off ?
(b) If the helicopter accelerates for 1.5
minutes, what velocity does it reach ?
Start by drawing a Force Diagram of the give
situation, including all Force Arrows. Identify
the given data.
Click to Advance
46
PROBLEM 9
A Helicopter weighs 2.0 x 103 N. Its rotating
blades exert 2.5 x 103 N. of force as it
lifts-off the ground. (a) What is the
helicopters acceleration during lift
off ? (b) If the helicopter accelerates for
1.5 minutes, calculate final velocity.
Wt 2.0 x 104 N FT 2.5 x 104 N
Fa
Fnet
wt
Determine the equations necessary to solve the
problem.
Click to Advance
47
PROBLEM 9
A Helicopter weighs 2.0 x 103 N. Its rotating
blades exert 2.5 x 103 N. of force as it
lifts-off the ground. (a) What is the
helicopters acceleration during lift
off ? (b) If the helicopter accelerates for
1.5 minutes, calculate final velocity.
Wt 2.0 x 104 N FT 2.5 x 104 N
Fa
a Fnet FT - wt m m
Fnet
Also v2 at v1
wt
Plug in the number and solve
Click to Advance
48
PROBLEM 9
A Helicopter weighs 2.0 x 103 N. Its
rotating blades exert 2.5 x 103 N. of force as
it lifts-off the ground. (a) What is the
helicopters acceleration during lift
off ? (b) If the helicopter accelerates for
1.5 minutes, calculate final velocity.
Wt 2.0 x 104 N FT 2.5 x 104 N
m wt/g 2.04 x 103 Kg
FT
a Fnet FT - wt m m
Fnet
a FT - wt 2.5 x 104 - 2.0 x 104
m 2.04 x 103 Kg a 2.45 m/s2
wt
Now solve for v2
Click to Advance
49
PROBLEM 9
A Helicopter weighs 2.0 x 103 N. Its
rotating blades exert 2.5 x 103 N. of force as
it lifts-off the ground. (a) What is the
helicopters acceleration during lift
off ? (b) If the helicopter accelerates for
1.5 minutes, calculate final velocity.
m wt/g 2.04 x 103 Kg
Wt 2.0 x 104 N FT 2.5 x 104 N
FT
a 2.45 m/s2
Fnet
v2 at v1 v2 (2.45 m/s2)(90 s) 221 m/s
wt
Click to Advance To The Next Problem
50
PROBLEM 10
A box having a mass of 50 kg is dragged
across a horizontal floor by means of a rope
tied on the front of it. The coefficient of
sliding friction between the box and the floor
is 0.30. If the angle between the rope and the
floor is 30o, what force must be exerted on the
rope to move the box at constant speed?
Start by drawing a Force Diagram of the give
situation, including all Force Arrows. Identify
the given data.
Click to Advance
51
PROBLEM 10
A box having a mass of 50 kg is dragged
across a horizontal floor by means of a rope
tied on the front of it. The coefficient of
sliding friction between the box and the floor
is 0.30. If the angle between the rope and the
floor is 30o, what force must be exerted on the
rope to move the box at constant speed?

• m 50 Kg
• µ .3
• 30o
• Fa ?

Fa
Fy
Ff
?
Fx
Fn
Now identify the appropriate formulas that apply
to this problem
Click to Advance
52
PROBLEM 10
A box having a mass of 50 kg is dragged
across a horizontal floor by means of a rope
tied on the front of it. The coefficient of
sliding friction between the box and the floor
is 0.30. If the angle between the rope and the
floor is 30o, what force must be exerted on the
rope to move the box at constant speed?
Fa
Fy

• m 50 Kg
• µ .3
• 30o
• Fa ?

Ff
?
Fx
Fn
Since this is uniform motion Fx Ff , also
Fn wt - Fy . Other useful formulas include
Fx cos?Fa and Fy sin?Fa and wt mg
Click to Advance
53
PROBLEM 10
A box having a mass of 50 kg is dragged
across a horizontal floor by means of a rope
tied on the front of it. The coefficient of
sliding friction between the box and the floor
is 0.30. If the angle between the rope and the
floor is 30o, what force must be exerted on the
rope to move the box at constant speed?
Fa
Fy

• m 50 Kg
• µ .3
• 30o
• Fa ?

Ff
?
Fx
Fn
This problem requires an algebraic solution.
Try to manipulate the equations from the
previous slide to solve the problem. Start with
Fx Ff
Click to Advance
54
PROBLEM 10
A box having a mass of 50 kg is dragged
across a horizontal floor by means of a rope tied
on the front of it. The coefficient of sliding
friction between the box and the floor is 0.30.
If the angle between the rope and the floor is
30o, what force must be exerted on the rope to
move the box at constant speed?
Fa
Fy

• m 50 Kg
• µ .3
• 30o
• Fa ?

Ff
?
Fx
Fn
Fx Ff cos30oFa µFn cos30oFa
µ(wt-Fy)
Distribute µ
Cos 30oFa µ(wt-sin 30oFa) µmgt - µsin
30oFa
Add µsin 30oFa
Cos 30oFa µsin 30oFa µmg
Factor out Fa
Fa(Cos 30o µsin 30o) µmg
Click to Advance
55
PROBLEM 10
A box having a mass of 50 kg is dragged
across a horizontal floor by means of a rope tied
on the front of it. The coefficient of sliding
friction between the box and the floor is 0.30.
If the angle between the rope and the floor is
30o, what force must be exerted on the rope to
move the box at constant speed?
Fa
Fy

• m 50 Kg
• µ .3
• 30o
• Fa ?

Ff
?
Fx
Fn
Fa(Cos 30o µsin 30o) µmg
Divide Both sides by Cos 30o µsin 30o
Fa µmg (Cos 30o
µsin 30o)
Plug in numbers and solve
Click to Advance
56
PROBLEM 10
A box having a mass of 50 kg is dragged
across a horizontal floor by means of a rope tied
on the front of it. The coefficient of sliding
friction between the box and the floor is 0.30.
If the angle between the rope and the floor is
30o, what force must be exerted on the rope to
move the box at constant speed?
Fa
Fy

• m 50 Kg
• µ .3
• 30o
• Fa ?

Ff
?
Fx
Fn
Fa (.3)(50 Kg)(9.81 m/s2) (Cos 30o
(.3)sin 30o)
Fa 147 N 144.6 N 1.016
Click to Advance To Next Problem
57
PROBLEM 11
A toboggan is sliding on snow at 8 m/s. How far
will the toboggan slide over the snow before
stopping if the coefficient of kinetic friction
is .03.
Start by drawing a Force Diagram of the give
situation, including all Force Arrows. Identify
the given data.
Click to Advance
58
PROBLEM 11
A toboggan is sliding on snow at 8
m/s. How far will the toboggan slide over the
snow before stopping if the coefficient of
kinetic friction is .03.
Ff
sled
Vx 8 m/s µ .03 V2 0 m/s
Determine the formulas v22 v12 2ad so
d v22 - v12 2a To solve this problem we
must know the force exerted on the sled to stop
it from sliding and then calculate the
acceleration.
0
Click to Advance
59
PROBLEM 11
A toboggan is sliding on snow at 8 m/s. How far
will the toboggan slide over the snow before
stopping if the coefficient of kinetic friction
is .03.
vx
Ff
sled
V1 8 m/s µ .03 V2 0 m/s
The force causing the octopus to slow (Fnet)
is friction. Ff µFn. To find
distance you first have to find acceleration. a
Fnet µFn µmg µg (.03)(9.81m/s2)
m m m .2943 m/s2
Click to Advance
60
PROBLEM 11
A toboggan is sliding on snow at 8 m/s. How far
will the toboggan slide over the snow before
stopping if the coefficient of kinetic friction
is .03.
vx
Ff
sled
Vx 8 m/s µ .03 V2 0 m/s a
-.2943 m/s2 (- since Fnet is opposite motion)
Now substitute a into the original
Equation d v22 - v12 2a
Click to Advance
61
PROBLEM 11
A toboggan is sliding on snow at 8 m/s. How far
will the toboggan slide over the snow before
stopping if the coefficient of kinetic friction
is .03.
vx
Ff
sled
Vx 8 m/s µ .03 V2 0 m/s a -.2943
m/s2 (- since Fnet is opposite motion)

Now substitute a into the original
Equation d - v12 2a
Click to Advance
62
PROBLEM 11
A toboggan is sliding on snow at 8 m/s. How
far will the toboggan slide over the snow before
stopping if the coefficient of kinetic friction
is .03.
vx
Ff
sled
Vx 8 m/s µ .03 V2 0 m/s a -.2943
m/s2 (- since Fnet is opposite motion)

d - v12 -(8 m/s)2 -64 m2/s2
2a 2(-.2943 m/s2) -.5886 m/s2
d 108.73 m
Click to Advance To The Next Problem
63
PROBLEM 12
A winch is used to pull a 1500 Kg boat up a
ramp that makes a 20o angle with the water at a
constant speed. The coefficient of friction
between the boat and ramp is .35 and the boat
moves a distance 25 m in 45 seconds. Calculate
the force generated by the winch.
Start by drawing a Force Diagram of the give
situation, including all Force Arrows. Identify
the given data.
Click to Advance
64
PROBLEM 12
A winch is used to pull a 1500 Kg boat up a
ramp that makes a 20o angle with the water at a
constant speed. The coefficient of friction
between the boat and ramp is .35 and the boat
moves a distance 25 m in 45 seconds. Calculate
the force generated by the winch.
Fa
Ff
F
20o
?
Fn
mg
F
Generate appropriate formulas to solve the
problem. Since Fa is being supplied by the
winch, that is what your looking for.
Click to Advance
65
PROBLEM 12
A winch is used to pull a 1500 Kg boat up a
ramp that makes a 20o angle with the water at a
constant speed. The coefficient of friction
between the boat and ramp is .35 and the boat
moves a distance 25 m in 45 seconds. Calculate
the force generated by the winch.
Fa
Ff
F
20o
?
Fn
mg
Fa F Ff (um) Ff µFn µcos?mg F
sin?mg
F
Do calculations !
Click to Advance
66
PROBLEM 12
A winch is used to pull a 1500 Kg boat up a
ramp that makes a 20o angle with the water at a
constant speed. The coefficient of friction
between the boat and ramp is .35 and the boat
moves a distance 25 m in 45 seconds. Calculate
the force generated by the winch.
Fa
Ff
F
20o
?
Fn
mg
Ff µFn µcos?mg .35(cos20o)(1500Kg)(9.81m
/s2) 4839.7 N F sin?mg
(sin20o)(1500Kg)(9.81m/s2) 5032.8 N Plug these
numbers into Fa F Ff (um)
F
Click to Advance
67
PROBLEM 12
A winch is used to pull a 1500 Kg boat up a
ramp that makes a 20o angle with the water at a
constant speed. The coefficient of friction
between the boat and ramp is .35 and the boat
moves a distance 25 m in 45 seconds. Calculate
the force generated by the winch.
Fa
Ff
F
20o
?
Fn
mg
Fa F Ff (um) 4839.7 N 5032.8 N
9872.53 N Fwinch 9872.53 N
F
Click to Advance To The Next Problem
68
PROBLEM 13
A store owner wishes to hang a sign weighing
275 N so that cable A attached to the store makes
a 20.0o angle as shown in the diagram. Cable B
is attached to an adjoining building. Calculate
the necessary tension (force) in cables A B.
Start by drawing a force diagram and identifying
and labeling all the given information.(The Data)
Next generate the necessary equations to solve
the problem.
Click to Advance
69
PROBLEM 13
A store owner wishes to hang a sign weighing
275 N so that cable A attached to the store makes
a 20.0o angle as shown in the diagram. Cable B
is attached to an adjoining building. Calculate
the necessary tension (force) in cables A B.
Fx
275 N
Since the sign is in equilibrium, The upward
and downward forces must be equal (275 N)
Likewise Fx And B are also equal. Generate
formulas to calculate Fx And thus B. Now solve
for the tension on cable A.
Fx
275 N
Click to Advance
70
PROBLEM 13
A store owner wishes to hang a sign weighing
275 N so that cable A attached to the store makes
a 20.0o angle as shown in the diagram. Cable B
is attached to an adjoining building. Calculate
the necessary tension (force) in cables A B.
Fx
275 N
Use the tangent function to Calculate
Fx. TanØ Opposite / Adjacent Tan70o 275 N /
Fx Fx 275 N 100.1 N Tan
70o Fx 100.1 N B
Fx
275 N
Click to Advance
71
PROBLEM 13
A store owner wishes to hang a sign weighing
275 N so that cable A attached to the store makes
a 20.0o angle as shown in the diagram. Cable B
is attached to an adjoining building. Calculate
the necessary tension (force) in cables A B.
Fx
275 N
To calculate A you need to use Sine or
cosine. SineØ Opposite / Hypotonuse Sine
70o 275 n / A A 275 N Sine 70o
A 292.65 N
Fx
275 N
Click to Advance To The Next Problem
72
PROBLEM 14
A 450 Kg block is attached over a
pulley to a 575 Kg mass as indicated in the
diagram below. The angle of incline is 35o and
the coefficient of friction is .57. Calculate all
the values indicated below

• Fa
• Wt
• Fn
• F
• Fnet
• Ff
• a

m2 450 Kg
575 Kg m1
35o
Start by drawing a force diagram and identifying
the given information. (The Data)
Click to Advance
73
PROBLEM 14
A 450 Kg block is attached over a
pulley to a 575 Kg mass as indicated in the
diagram below. The angle of incline is 35o and
the coefficient of friction is .57. Calculate all
the values indicated below
Fa
Fnet

• Fa
• Wt
• Fn
• F
• Fnet
• Ff
• a

m2 450 Kg
575 Kg m1
Ff
F
?
Fn
mg
?
wt

• 35o
• µ .57

F
Generate the necessary equations to solve for
the values requested.
Click to Advance
74
PROBLEM 14
A 450 Kg block is attached over a
pulley to a 575 Kg mass as indicated in the
diagram below. The angle of incline is 35o and
the coefficient of friction is .57. Calculate all
the values indicated below
Fa
Fnet

• Fa
• Wt
• Fn
• F
• Fnet
• Ff
• a

m2 450 Kg
575 Kg m1
Ff
F
?
Fn
mg
?
wt

• 35o
• µ .57

Fa wt Wt m1g Fn cos?m2g Fnet Fa
- (Ff F) F sin?m2g a Fnet Ff
µFn msystem Plug in numbers and
calculate.
F
Click to Advance
75
PROBLEM 14
A 450 Kg block is attached over a
pulley to a 575 Kg mass as indicated in the
diagram below. The angle of incline is 35o and
the coefficient of friction is .57. Calculate all
the values indicated below
Fa
Fnet

• Fa
• Wt
• Fn
• F
• Fnet
• Ff
• a

m2 450 Kg
575 Kg m1
Ff
F
?
Fn
mg
?
wt

• 35o
• µ .57

F
Wt m1g (575 Kg)(9.81 m/s2) 5640.75 N
Fa Fn cos?m2g cos35o(450 Kg)(9.81m/s2)
3616.15 N F sin?m2g sin35o(450
Kg)(9.81m/s2) 2532.05 N Now Calculate Fnet
Click to Advance
76
PROBLEM 14
A 450 Kg block is attached over a
pulley to a 575 Kg mass as indicated in the
diagram below. The angle of incline is 35o and
the coefficient of friction is .57. Calculate all
the values indicated below
Fa
Fnet

• Fa
• Wt
• Fn
• F
• Fnet
• Ff
• a

m2 450 Kg
575 Kg m1
Ff
F
?
Fn
mg
?
wt

• 35o
• µ .57

Fa 5640.75 N Fn 3616.15 N F 2532.05 N
Ff 2061.21 N Fnet Fa - (Ff F) 5640.75
N - (2061.21 N 2532.05 N) 5640.75 N -
(4593.26 N) 1047.49 N a Fnet
msystem Plug in numbers and calculate.
F
Click to Advance
77
PROBLEM 14
A 450 Kg block is attached over a
pulley to a 575 Kg mass as indicated in the
diagram below. The angle of incline is 35o and
the coefficient of friction is .57. Calculate all
the values indicated below
Fa
Fnet

• Fa
• Wt
• Fn
• F
• Fnet
• Ff
• a

m2 450 Kg
575 Kg m1
Ff
F
?
Fn
mg
?
wt

• 35o
• µ .57

Fa 5640.75 N Fn 3616.15 N F 2532.05 N
Ff 2061.21 N a Fnet Fnet 1047.49 N
msystem Now Calculate
Acceleration
F
Click to Advance
78
PROBLEM 14
A 450 Kg block is attached over a
pulley to a 575 Kg mass as indicated in the
diagram below. The angle of incline is 35o and
the coefficient of friction is .57. Calculate all
the values indicated below
Fa
Fnet

• Fa
• Wt
• Fn
• F
• Fnet
• Ff
• a

m2 450 Kg
575 Kg m1
Ff
F
?
Fn
mg
?
wt

• 35o
• µ .57

Fa 5640.75 N Fn 3616.15 N F 2532.05 N
Ff 2061.21 N a 1047.49 N Fnet
1047.49 N 1025 Kg a
1.02 m/s2
F
Click to Advance To Next Slide
79
PROBLEM 15
Calculate the net force required to accelerate a
20 Kg. lawn mower across the yard at 2 m/s2, if
a force is applied to the handle at a 60o angle
above the horizontal, and the coefficient of
friction is .8. How much force would have to be
applies to the handle.
Start by drawing a Force Diagram of the give
situation, including all Force Arrows. Identify
the given data.
Click to Advance
80
PROBLEM 15
Calculate the net force required to accelerate a
20 Kg. lawn mower across the yard at 2 m/s2, if
a force is applied to the handle at a 60o angle
above the horizontal, and the coefficient of
friction is .8. How much force would have to be
applies to the handle.
Fa
Fy
a 2 m/s2 M 20 Kg µ .8 ? 60o
Ff
Fx
Wt mg
Generate appropriate formula for Fnet. Plug in
values and solve.
Click to Advance
81
PROBLEM 15
Calculate the net force required to accelerate a
20 Kg. lawn mower across the yard at 2 m/s2.
Calculate the coefficient of friction if a force
required is 100 N applied to the handle at a 60o
angle above the horizontal.
Fa
Fy
a 2 m/s2 M 20 Kg Fa 100 N ? 60o
Ff
?
Fx
Wt mg
Fnet ma (20 Kg)(2 m/s2) 40 N To find µ
you need to find Ff and Fn. Generate formulas to
solve.
Click to Advance
82
PROBLEM 15
Calculate the net force required to accelerate a
20 Kg. lawn mower across the yard at 2 m/s2.
Calculate the coefficient of friction if a force
required is 100 N applied to the handle at a 60o
angle above the horizontal.
Fa
Fy

• a 2 m/s2
• M 20 Kg
• Fa 100 N
• 60o
• Fnet 40 N

Ff
?
Fx
Wt mg
µ Ff Fn mg - Fy Fn
Fy sin?Fa Ff can be calculated from Fnet
and Fx Generate the appropriate equation.
Click to Advance
83
PROBLEM 15
Calculate the net force required to accelerate a
20 Kg. lawn mower across the yard at 2 m/s2.
Calculate the coefficient of friction if a force
required is 100 N applied to the handle at a 60o
angle above the horizontal.
Fa
Fy

• a 2 m/s2
• M 20 Kg
• Fa 100 N
• 60o
• Fnet 40 N

Ff
?
Fx
Wt mg
µ Ff Fn mg - Fy Fn
Fy sin?Fa Fnet Fx - Ff So Ff Fx - Fnet
and Fx cos?Fa On to the calculations.
Click to Advance
84
PROBLEM 15
Calculate the net force required to accelerate a
20 Kg. lawn mower across the yard at 2 m/s2.
Calculate the coefficient of friction if a force
required is 100 N applied to the handle at a 60o
angle above the horizontal.
Fa
Fy

• a 2 m/s2
• M 20 Kg
• Fa 100 N
• 60o
• Fnet 40 N

Ff
?
Fx
Wt mg
Fy sin?Fa sin60o(100 N) 86.60 N Fn
mg - Fy (20 Kg)(9.81 m/s2) - (86.60 N) 109.6
N Fx cos?Fa cos60o(100 N) 50.00 N Fnet
Fx - Ff So Ff Fx - Fnet 50 N - 40 N
10 N µ Ff 10 N .091
Fn 109.6 N
Click to Advance
85
Suggestions
Look over all the tests we took this
semester. Review all vocabulary. If you can do
the problem in this review you should be well
prepared for the exam. I Hope This Was a Helpful
Review !!! GOOD LUCK !!!!!!!!