NAME: ________________________________ DATE: ____________ PERIOD: ______
MR. TALBOO – PHYSICS
1st SEMESTER REVIEW
PRACTICE PROBLEM A
You and your friend are pulling a broken down Hummer. You and your friend attach ropes and pull the
vehicle with the forces shown in the diagram below. A 3500 N frictional force is resisting the motion of
the vehicle.
2500 Newtons
3500 Newtons
20°
35°
2200 Newtons
(a) What is the resultant force acting on the vehicle?
(magnitude and direction)
Ax = -3500 N
Ay = 0 N
Bx = B cos θ = 2500 cos 20 = 2349.232 N
By = B sin θ = 2500 sin 20 = 855.050 N
Cx = C cos θ = 2200 cos 35 = 1802.134 N
Cy = C sin θ = 2200 sin 35 = -1261.868 N
Rx = Ax + Bx + Cx = -3500 + 2349.232 + 1802.134 = 651.366 N
Ry = Ay + By + CY = 0 + 855.050 – 1261.868 = -406.818 N
R = √(Rx2 + Ry2) = √(651.3662 + 406.8182) = 767.970 N
θ = tan-1 (406.818 / 651.366) = 31.987°
R = 767.970 N @ 31.987° S of E
(b) If the H2 has a mass of 2994 kg, what is the horizontal acceleration of the vehicle?
ax = ΣFx / m = 651.366 / 2994 = 0.218 m/s2
PRACTICE PROBLEM B
Create a position and velocity vs. time graph for each description below:
1. ∆x = positive
S = increasing
a = positive
x
2. ∆x = negative
S = increasing
a = negative
x
v
t
t
v
t
3. ∆x = positive
S = decreasing
a = negative
4. ∆x = positive
S = constant (non-zero)
a = zero
5. ∆x = negative
S = constant (non-zero)
a = zero
6. ∆x = negative
S = decreasing
a = positive
x
t
v
t
x
t
v
t
x
t
v
t
x
t
v
t
t
PRACTICE PROBLEM C
Starting from rest, a top fuel dragster reaches a velocity of 100 mi/hr in a time of 1.2 s.
(A) Calculate the car’s average acceleration. (B) Calculate the car’s displacement during this
time interval.
∆x
?
DATA TABLE
vi
vf
a
0
44.694
37.245
t
1.2
A. a = (Vf – Vi) / t = (44.694 – 0) / 1.2
= 37.245 m/s2
B. ∆x = Vit + ½ at2 = 0.5(37.245)1.22 = 26.816 m
PRACTICE PROBLEM D
A man’s 2012 BMW goes from 60 mi/hr to 20 mi/hr with a
constant acceleration of – 21.032 m/s2. (A) How long does it
take the car to slow to this speed? (B) Find the distance the car
travels when slowing down.
DATA TABLE
∆x
vi
vf
a
t
-
26.817
8.939
-21.032
?
(A) t = (Vf – Vi) / a = (8.939 – 26.817 ) / -21.032 = 0.850 s
(B) ∆x = (Vf2 – Vi2) / 2a = (8.9392 – 26.8172) / (2(-21.032)) = 15.197 m
PRACTICE PROBLEM E
A golf ball is thrown into the air with a velocity of 10 m/s while you are standing on top of a
building 30 m high. The ball misses the roof, and falls to the ground. Determine (a) the time the
golf ball is at its maximum height, (b) the maximum height, (c) the total time it is in the air, and
(d) the velocity of the golf ball when it hits the ground.
a) t = Vf - Vi / g = (0 – 10) / -9.8 = 1.020 s
b) ∆y = (Vf2 – Vi2) / 2g = (02 – 102) / (2 x -9.8) = 5.102 m
+ 30 m
Max height = 35.102 m
c) t = √(2∆y/g) = √(2 x -35.102 / -9.8) = 2.676 s
total time = 2.676s + 1.020 s = 3.696 s
d) Vf = Vi + gt = 0 + (-9.8 x 2.676) = -26.224 m/s
PRACTICE PROBLEM F
A plane drops a package of supplies to a party of explorers. If the plane is traveling horizontally
at 49.0 m/s and is 1200 m above the ground, where does the package strike the ground relative to
the point at which it is released?
∆y = ½ gt2
t = √(-2 ∆y/g) = √(-2 (-1200)/9.8) = 15.649 s
∆x = Vx t = (49 m/s) (15.649 s) = 766.801 m
PRACTICE PROBLEM G
A long-jumper leaves the ground at an angle of 24° above the horizontal and at a speed of 11.6
m/s.
(a) How long is he in the air?
(b) How far does he jump in the horizontal direction?
(c) What is his maximum height?
A) Vxi = 11.6 cos 24° = 10.597 m/s
Vyi = 11.6 sin 24° = 4.718 m/s
t = (Vf – Vi) / -g = (-4.718 – 4.718) / -9.8 = 0.963 s
B) R = (Vi2 sin 2θ) / g = (11.62 sin 48) / 9.8 = 10.204 m
C) H = (Vi2 sin2θ) / 2g = (11.62 (sin 24)2) / 2(9.8) = 1.136 m
PRACTICE PROBLEM H
An 80kg box slides down an inclined plane that makes an angle of 26°
with the horizontal. If a frictional force of 266 N is applied to the box as
it slides down the incline, (a) find the acceleration of the box. (b) If the
box slides for a displacement of 30.0 m, what is the final velocity of the
box?
Forces
x-components
y-components
Fg
343.683
-704.654
Ffr
-266
0
Fn
0
704.654
ΣFx = 77.683 N
m
θ
Ffr
Fn
ΣFy = 0 N
ax = ΣFx / m = (77.683 / 80) = 0.971 m/s2
ay = 0 m/s2
Vf = √(Vi2 + 2a∆x) = √(02 + 2(0.971)(30)) = 7.633 m/s
Fg
PRACTICE PROBLEM I
Canadian Kevin Fast pulled a 37,000 kg airplane a distance of 20 m. If Kevin pulled the plane at
an angle of 15° with a force of 267000 N while friction
resisted its motion with a force of 254400 N,
(a) what was the acceleration of the plane?
(b) what was the plane’s final velocity?
Forces
x-components
y-components
FT
257902.196
69104.685
Fg
0
-362600
Fn
0
293495.315
Ffr
-254400
0
Fn
ΣFx = 3502.196 N ΣFy = 0 N
ax = ΣFx / m = (3502.196 / 37000) = 0.095 m/s2
ay = 0 m/s
Ffr
FT
2
Fg
Vf = √(Vi2 + 2a∆x) = √(02 + 2(0.095)(20)) = 1.949 m/s