Practice
A plane flies 1500 miles East and 200 miles South. What is the magnitude and
direction of the plane’s final displacement?
A hiker walks 80 m North, 20 m East, 50 m South, and 30 m West. What is the
magnitude and direction of the hiker’s displacement?
A plane flies 1500 miles East and 200 miles South. What is the magnitude and direction
of the plane’s final displacement?
**not drawn to scale**
1500 miles
θ
200
miles
a2 + b2 = c2
(1500 m)2 + (200 m)2 = R2
R = √ (1500 m)2 + (200 m)2
R = 1513.275 m
tan θ = opp/adj
θ = tan-1 (200/1500)
θ = 7.5946433°
A hiker walks 80 m North, 20 m East, 50 m South, and 30 m West. What is the
magnitude and direction of the hiker’s displacement?
By subtracting the opposing directions from each
other, we find the hiker’s displacement along the y-axis
to be 30 m North, and the displacement on the x-axis
to be 10 m West.
a2 + b2 = c2
302 + 102 = R2
R = √900 + 100
R = 31.623 m
tan θ = opp/adj
θ = tan-1 (10/30)
θ = 18.435°
Practice Problem #1
An airplane traveling 1001 m above the ocean at 125 km/h
is going to drop a box of supplies to shipwrecked victims
below.
a. How many seconds before the plane is directly
overhead should the box be dropped?
b. What is the horizontal distance between the
plane and the victims when the box is dropped?








Practice Answer #1
Vertical
viy = 0 m/s
ay = -9.8 m/s2
dy = 1001 m
t=?
d = ½ at2
1001 m = ½ (9.8 m/s2)t2
t = 14.3 s
Horizontal
vix = 125 km/h
dx = ?
t=?
ax = 0 m/s2
dx = vixt
dx = (125 km/h)(14.3 s)
dx = (0.03472 km/s)(14.3 s)
dx = 0.4965 km = 496.5 m
Practice Problem #2
Herman the human cannonball is launched
from level ground at an angle of 30° above
the horizontal with an initial velocity of 26
m/s. How far does Herman travel horizontally
before reuniting with the ground?
Practice answer
#2
Horizontal
 Vertical
 viy = 26(sin30)
 ay = -9.8 m/s2
 vf = 0 m/s
 t=?
 vf = viy + at
 0 = 26(sin30) + (-9.8 m/s2)t
 -13 = -9.8 m/s2t
 t = 1.33 s
vix = 26(cos30)
ax = 0 m/s2
dx = ?
t=?
dx = vixt
dx = 26(cos30)(2.66 s)
dx = 60.0 m
Case 2: Block on an inclined plane
FNormal
Fperpendicular(F|) = Fgcosθ
FN = F |
FParallel
FPerpendicular
FGravity
Fparallel (F||) causes sliding
F|| = Fgsinθ
Practice Problem 1 (part 1)
A 50 kg sled is pulled by a boy across a smooth, icy
surface. If the boy is pulling the sled 500 N at 30° above
the horizontal, what is the horizontal component of the
force?
30°
sled
Practice Problem 1 (part 2)
What is the acceleration experienced by the
sled?
Does the vertical component of the force affect
the acceleration?
sled
Answers
Ax = Acosθ
Fx = (500 N)cos(30°)
Fx = 430 N
FNET = ma
430 N = (50 kg)a
a = 8.6 m/s2
The vertical component doesn’t affect
acceleration. It only causes the object to lift off the
ground, rather than move it backward or forward.
Practice Problem 1 (part 3)
Using the same sled from the previous example, what
is the weight of the sled?
Find the vertical component of the force pulling the
sled. Would this force cause the sled to lift of the
ground? Why?
sled
Practice Problem 1 (part 4)
What is the Normal Force felt by the sled?
sled
Answers
Weight is equal to Fg. Fg = mg. Since the mass of the
sled is 50 kg, we can find weight by plugging in the
numbers and solving for Fg.
Fg = (50 kg)(9.81 m/s2)
Fg = 490 N down
Ay = Asinθ
Fy = 500 N sin(30°)
Fy = 250 N up
This would not be enough to lift
the sled because the force due to
gravity is much greater.
Answers
The Normal Force felt by the sled would be equal in
magnitude to the Gravitational Force (Weight), but in
the opposite direction (perpendicular to the surface).
Therefore, FN = 490 N up
Practice Problem 2 (part 1)
A block with a mass of 100 kg is at rest on an inclined
plane with an angle of 30°. What is the weight of the
block?
What is the parallel force of the block?
What is the perpendicular force of the block?
30°
Drawing a Force Diagram
Green Vector represents the force
due to Gravity (Weight = mg)
Red Vector represents the
Normal Force of the incline
pushing up on the box
30°
Blue Vectors represent the components of the weight (Perpendicular and
Parallel Forces)
Answers
The Weight of the block is equal to mass times acceleration due to gravity.
W = mg
W = (100 kg)(9.8 m/s2)
W = 980 N
F|| = Fg sinθ
F|| = (980 N)sin(30°)
F|| = 490 N
F| = Fg cosθ
F| = (980 N)cos(30°)
F| = 850 N
Practice Problem 2 (part 2)
What is the acceleration of the block as it slides down
the inclined plane?
What is the Normal Force felt by the block?
As the angle of the inclined plane increases, what
happens to the parallel and perpendicular forces?
30°
Answers
To find the acceleration of the block as it slides down the
incline, we need to use the parallel force.
F|| = ma
490 N = (100 kg)a
a = 4.9 m/s2
Normal Force is equal to the Perpendicular Force, but opposite
in direction.
FN = 850 N
Related Problems
Energy Conservation Practice
Energy Conservation Practice
Energy Conservation Practice
Energy Conservation Practice
Energy Conservation Practice
Energy Conservation Practice