PROBLEM 1
• A stone is thrown from a hill at an angle
of 600 with the horizontal at an initial
velocity of 30 m/sec. After hitting level
ground at the base of the hill, the stone
has covered a horizontal distance of 150
m. How high is the hill?
A. 210.7m
B. 220.7m
C. 230.7m
D. 240.7m
SOLUTION
X= vcosφ.t
150=30cos60.t
T= 10 sec
Y=vsin φ.t – ½ (gt)2
Y=-h
-h = 30sin60o(10)-1/2 (9.81)(10)2
-h = -230.7 m
H= 230.7 m
PROBLEM 2
• A shell leaves a mortar with a muzzle
velocity of 150 m/sec directed
upward at 600 with the horizontal.
Determine the resultant velocity 20
sec after firing. How will it rise?
A. V = 100m/sec ; h = 860m
B. V= 110m/sec ; h = 870m
C. V = 200m/sec ; h = 960m
D. V = 210m/sec; h = 970m
SOLUTION
X= Vcos602.t
X= 150cos60o(20)
X= 1500m
Y =vsin φ.t – ½ (gt)2
Y = 150sin60o(10)-1/2 (9.81)(20)2
Y= 636m
Vbx=Vax
Vbx=150cos60o
Vbx=75m/s
Vby2 = (150sin60o)2 - 2(9.81)(636)
Vby= 66.31m/s
Vby=√ Vbx2 +Vby2
V= √(75)2+(68.31)2
V = 100m/s
H = V2sin2φ/2g
H= (150)2(sin60)2
2(9.81)
H = 860 m
PROBLEM 3
• A projectile is fired with the initial
velocity of 60m/sec upward at an angle
of 30o to the horizontal from a point 80m
above a level plain. What horizontal
distance will it cover before it strikes the
plane?
A. 534m
B. 422m
C. 638m
D. 345m
SOLUTION
Y = xtanφ – gx2/2V2cos2φ
-80= xtan3oo – 9.81x2/2(60)2cos230o
-80= 0.774x – 0.001817x2
0.001817x2-0.5774 – 80 = 0
X=
X = 422 m
PROBLEM 4
• The car shown is just to clear the
water filled gap. Find the take off
velocity.
A. 4.64m/sec
B. 3.63m/sec
C. 6.44m/sec
D. 7.63m/sec
SOLUTION
Y = xtanφ-6 = 5.5tan30o—
-9.175= V2 = 21.56
V= 4.64m/sec
PROBLEM 5
• A stunt man is to drive a car across
the water filled gap. Determine
the car’s minimum take- off
velocity and angle φ of the landing
ramp.
A.90o
B.1800
C.30o
D.45o
SOLUTION
Y = xtanϕ-3 = 12(1/2) —
9= V2 =98.1
V= 9.9m/s
Vay2= vcosϕ = 8.855m/sec
Vay= Vay2- 2gh = (9.90(1/√5))2 – 2 (9.81)(-3)
Vay=8.858m/s
Tanϕ=
=
Φ= 45o
PROBLEM 6
• In the fig , a ball thrown down the incline
strikes it at a distance S = 75 m. if the ball
rises to a maximum height h = 20m. above
the point of release , compute the initial
velocity and the inclination φ.
A. 14.4m/sec
B. 44.4m/sec
C. 24.4m/sec
D. 34.4m/sec
SOLUTION
= =
Y= 23.72m
X= 71.15m
H = V2sin2φ/2g
20 =
20 =
V2sin2φ=392.4
V2 =
Y = xtanϕ -23.72 = 71.15 tanϕTanϕ =
Φ= 54.3o
V2 =
= 24.4 m/s
= 71.15tanϕ – 63.28 tan 2ϕ
=
= 1.39338
PROBLEM 7
• A rocket is released at a jet fighter flying
horizontally at 1200 kph at an altitude of
2400 m. above its target the rocket thrust
gives its constant horizontal acceleration of
0.6g. determine the angle between the
horizontal and the line of sight to the target.
A. 15.23o
B. 34.43o
C. 45.27o
D.23.15o
SOLUTION
S=
2400 =
T = 22.12 sec
X = Vt +
X = 1200(
X = 8813 m
Tanϕ = 2400 / 8813
Tanϕ= 15.230
PROBLEM 8
• If the initial velocity of an object is
12m/sec, determine the horizontal
distance it can cover without
rising more than 3m.
A. 14.43m
B. 43.14m
C. 41.34m
D. 34.41m
SOLUTION
Φ = 39.74o
X=
X=
X = 14.43 m
PROBLEM 9
• A ball is thrown so that it can just clear
a 3m fence 18m away. If it left the
hand 1.5m above the ground at an
angle of 60o with the horizontal , what
was the initial velocity of the ball?
A. 64.14m/sec
B. 46.41m/sec
C. 16.64m/sec
D. 41.46m/sec
SOLUTION
Y = xtanϕ 1.5 = 18tan60o –
-29.677 = 6356.88 / v2
V2 = 6356.88 / 26.677
V = 14.64 m
PROBLEM 10
• the fig shown, find φ to cause the
projectile to hit B in exactly 4 sec.
what is the distance x?
A. 192.3m
B. 109.8m
C. 128.5m
D. 178.4m
SOLUTION
y = Vsinϕt +
-30= 30 sinϕ(4) +
-30= 120 sinϕ- 78.48
Φ = 23.83o
X = vcosϕt
X = 30(cos23.83o)(4)
X= 109.8m
PROBLEM 11
• Boat A moves with a constant velocity of
6m/sec. starting from the position shown.
Find φ in order for the projectile to hit the
boat 5 sec after starting, under the conditions
given. How high is the hill above the water?
A. Φ = 63.78o ; y = 23.625 m
B. Φ = 38.87o ; y = 32.625 m
C. Φ = 42.64o ; y = 54.625 m
D. Φ = 24.48o ; y = 45.625 m
SOLUTION
Y = Vsin ϕt
120 = (30cosϕ)(5)
Cosϕ = 120/150
Φ= 36.87O
y = Vsinϕt +
- y = 30(sin36.87)(5) +
-y = -32.625 m
Y = 32.625 m
PROBLEM 12
• It is desired to pitch a golf ball across a trap to a green
30m away what is the best club to useif the initial
velocity of the ball is 18m/sec. assume the ball stops
dead after striking the green, which is on the same
level as the point from which the ball is struck.
Assume the clubs have the slopes graduated at
intervals of 6o so that a number 1 iron has a face
inclined at 80o to the ground, the number 2 iron at
74o, down to a number 9 inclined at 32.5o
A. 54.2o
B. 66.4o
C. 95.3o
D. 32.5o
SOLUTION
R=
30 =
2ϕ= 65o
Φ = 32.5o
Use a number 9 iron
PROBLEM 13
• A boy running a foot race rounds a
flat curve of 15 m radius. If he runs
at the rate of 6.50 m/sec at what
angle with the vertical will he incline
his body?
A. 12o
B. 16o
C. 18o
D. 22o
SOLUTION
Tan ϕ = gr/w = v2/gr = (6.50)2/9.81(15)
Φ =16o
PROBLEM 14
• A daredevil drives a motorcycle around a
circular vertical wall 30m in diameter. The
coefficient of friction between the tires and
wall is 0.60 . what is the minimum speed that
will prevent sliding down the wall? At what
angle will the motorcycle be inclined with the
horizontal?
A. V = 56.4kph ; φ= 31o
B. V = 65.4kph ; φ= 32o
C. V = 73.4kph ; φ= 33o
A. V = 81.4kph ; φ= 34o
SOLUTION
N = wv2 / gr
F = µN = 0.6
=0
W = 0.6
V2 = gr/0.6 = (9.81)(15) / 0.60
V = 15.66 m/s 3600 / 1000 = 56.4 kph
Tan ϕ = 0.6
Φ = 31o this angle is with the wall
Φ = 59.04 angle with the wall
PROBLEM 15
• An airplane makes a turn in a horizontal
plane without sideslip at 774kph . at
what angle must the plane be banked if
the radius of turn is 1.75 km? of the pilot
weighs 600 N, what pressure does he
exert on his seat?
A. 2171.3 N, 80o
B. 1721.3 N, 69.6o
C. 3712.3 N, 12o
D. 7123.1 N, 96.9o
SOLUTION
Tan ϕ = v2 / gr
V = 774 kph 1000/3600
V = 215 m /s
Tan ϕ =
Φ = 69.6 0
Cosϕ = 600 /r
R = 1721.3 N
PROBLEM 16
• Find the angle of banking for a highway
curve of 90m radius designed to
accommodate cars traveling at 160kph, if
the coefficient of friction between the
tires and the road is 0.60. what is the
rated speed of the curve?
A. Φ= 34.9o; v= 89.4 kph
B. Φ= 43.9o; v= 98.4 kph
C. Φ= 93.4o; v= 43.4 kph
D. Φ= 73.4o; v= 75.4 kph
SOLUTION
a)Tan o = 0.60
O = 31o
V = 160 kph 1000/ 3600 = 44.44 m/s
Tan (a + o) = v2 / gr
Tan (a + 31) = (44.44)2 / 9.81(90)
A + 31 = 65.0 = 34.90
b)tan o = v2 / gr
tan 34.9o = V2 / 9.81(90)
v= 24.82 m/s 3600 / 1000 = 89.4 kph
PROBLEM 17
• The rated speed of a highway curve of
60m radius is 50 kph. If the coefficient of
friction between the tires and the road is
0.60, what is the maximum speed at
which the car can round the curve
without skidding?
A. 96.3 kph
B. 36.9 kph
C. 93.6 kph
D. 69.3 kph
SOLUTION
V = 50 kph = 50 ( 1000/3600) = 13.89 m / s 2
Tan a = v2/ gr = (13.89)2 / 9.81(60)
a = 18.15o
Tan (a + o) = v2 / gr
Tan o = 0.60
O = 31o
Tan (18.15 + 31) = V2 / 9.81(60)
V = 26 m/s 3600/ 1000 = 93.8 kph (max speed)
PROBLEM 18
• A 3000 N block slides down an incline
having a slope of 3 vertical to 4 horizontal. It
starts from rest and after moving 1.2m
strikes a spring whose modulus is 1500N/m.
if the coefficient of kinetic friction is 0.20,
find the maximum velocity of the block.
A. 3.23m/sec
B.2.32m/sec
C. 4.23m/sec
D. 1.23m/sec
SOLUTION
N = 4/5 ( 3000) = 2400 N
F = 2400(0.20) = 480N
1800(1.2) – 480(1.2) – 1500 (s)2 = ½ (w/g) (02-02)
S= 1.03m
Ks2/2 = ½ (w/g)(vmax2)
v max = 2.35m/s
PROBLEM 19
• A train weighs 16000 KN. The train
resistance is constant at 96000N. if
train up at 2 % grade. What will be
its speed in kph?
A. 38.7 kph
B. 73.8 kph
C. 83.7 kph
D. 67.8 kph
SOLUTION
P = 96000 + 16000(1000)(2/100) = 416000 N
Hp = PV / 746
6000 = 416000 V / 746
V = 10.76 m/sec (1 km / 1000 m) (3600 s / hr )
= 38.7 kph
PROBLEM 20
• A train weighing 1000 KN is being pulled
up to a 2 % grade. The train resistance is
5000 N. the speed of the train is increased
from 6 m/s to 12m/s. in a distance of 300
m. find the maximum horsepower
developed by the locomotive.
A. 349 hp
B. 230 hp
C. 697 hp
E. 956 hp
SOLUTION
P(300)= 5000(300) – 1000 (1000)(2/100)(300) =
½ (1000)(1000) / 9.81 ((12)2- (6)2)
P = 43349 N
Hp = pv / 746
Hp = 43349(12) / 746
Hp = 697 hp
PROBLEM 21
• A horizontal force of 1500 N pushes
1000N block up an incline whose slope
is 3 vertical 4 horizontal. If u k= 0.2 ,
determine the time required to increase
the velocity f the block from 3 to 15 m/s
A. 4.23 s
B. 2.45 s
C. 0.34 s
D. 1.65 s
SOLUTION
N = 1000 (4/5) = 800N
F= 0.20 (800) = 160 N
= w/g (V2-V1)
t = 1000/9.81 (15 – 3)
740t = 1000/9.81 (12)
T = 1.65 sec
PROBLEM 22
• A 5kN shell is fired from a 1000kN
cannon with a velocity 0f 600 m/s.
find the modulus of a nest of springs
that will limit the recoil of the
cannon to 0.90 m.
A. 1133kN/m
B. 5423kN/m
C. 3493kN/m
D. 4530kN/m
SOLUTION
M1v1=m2v2
5/g (600) = 1000/g (v2)
V2 = 3 m/s
½ ks2 = ½ w/g V2 2
1/2k(0.9)2 = ½ 1000/9.81 (3)2
K = 1133kN/m
PROBLEM 23
• A 6000 N hammer falling freely through
0.9m drives a 3000 N pile 150mm
vertically into the ground. Assuming the
hammer and the pile to cling together
after impact, determine the average
resistance to penetration of the pile.
A. 93000 N
B. 33000 N
C. 54000 N
D. 86000 N
SOLUTION
S=
0.15 =
0.15 =
R = 33000 N
PROBLEM 24
• A thin heavy uniform iron rod 16 m long is
bent at the 10 m mark forming a right angled
L- shaped pieced 6 m by 10m . the bend rod
is hung on a peg at the point of bend. What
angle does the 10m side make with the
vertical when the system is in equilibrium?
A. 19o48’
B. 28o12’
C. 24o36’
D. 26o14’
SOLUTION
6w(3)sin(90 -0) = 10w(5)sin0
18 sin(90-0) = 50sin 0
18(sin90cos0 – sin0cos90) = 50sin0
Tan0 = 1815
O = 19048’
PROBLEM 25
• A painters scaffold 30m long and a mass of
300kg is supported in a horizontal position by
vertical ropes attached at equal distances
from the ends of the scaffold. Find the
greatest distance from the ends that the
ropes maybe attached sp as to permit a 200kg
man to stand safely at one end of the scaffold
A. 9m
B. 8m
C. 6m
D. 7m
SOLUTION
200x = 300(30-2x /2 )
400x= 9000 – 600x
X = 9m
PROBLEM 26
• A cylindrical tank having a diameter of 16cm
weighing 100kN is resting on a horizontal floor. A
block having a height of 4cm is placed on the side
of the cylindrical tank to prevent it from rolling.
What horizontal force must be applied at the top
of the cylindrical tank so that it will start to roll
over the block? Assume the block will not slide
and firmly attached to the horizontal floor
A. 57.74kN
B. 68.36kN
C. 48.52kN
D. 75.42kN
SOLUTION
Cos 0 = 4/8
O = 60
X = 8sin60
X = 6.93
100(6.93) = p12
P = 57.74 kN
PROBLEM 27
• When one boy is sitting 1.20 m from the center of
the seesaw, another boy must sit on the other
side 1.50m from the center to maintain at even
balance . however, when the first boy carries an
additional weight of 14 kg and sit 1.8 m from the
center, the second boy must move to 3m from
the center of balance. Neglecting the weight of
the seesaw, find the weight of the heaviest boy
A. 58 kg
B. 42kg
C. 35kg
D. 29 KG
SOLUTION
Wa(1.20) = Wb (1.50)
Wa = 1.25Wb
Wa (1.8) + 14(1.8) = Wb(3)
(1.25Wb)(1.8) + 14(1.8) = 3Wb
0.75Wb = 14(1.8)
Wb = 33.6 kg
Wa = 1.25(33.6)
Wa = 42 kg
PROBLEM 28
• A man can exert a maximum pull of
1000N but wishes to lift a new stone
door for his cave weighing 20000 N. if he
uses lever , how much closer must the
fulcrum be to the stone than to his hand
A. 10 times nearer
B. 20 times farther
C. 10 times farther
D. 20 times nearer
SOLUTION
1000y= 20000x
Y = 20x
X= y/20
The stone is 20 times nearer
PROBLEM 29
• A block weighing 400kg is placed on an
inclined plane making an angle of φ from
the horizontal. If the coefficient of friction
between the block and the inclined plane
is 0.30. find the value of φ, when the
block impends to slide downward.
A. 13.60o
B. 16.70o
C. 15.80o
D. 14.50o
SOLUTION
W = cos o = N
F = µN = wsino
µN = w sin o
µW cos o= wsin o
tan o = µ = o.30
o = 16.70
PROBLEM 30
• The pull required to overcome the rolling
resistance of a wheel is 90 N acting at the
center of the wheels. If the weight of the
wheels is 18000 N and the diameter of
the wheels is 30 mm, determine the
coefficient of rolling resistance.
A.0.75mm
B.0.60mm
C.0.50mm
D.0.45mm
SOLUTION
18000x = 90(150)
X = 0.75 mm
PROBLEM 31
• A cable 800m long weighing 1.5
kg/m has tension of 750 kg at its
ends. Compute the sag of the
cable.
A.200m
B.100m
D.150m
C.50m
SOLUTION
T= wy
750 = 1.5y
Y = 55 m
2s = 800
S = 400 m
C = 300 m
D = y-c
D = 500 – 300 = 200 m
PROBLEM 32
• A racing car during the Marlboro
championship starts from rest and
has a constant acceleration of 4m/s2.
What is its average velocity during
the first 5 sec of motion?
A. 10m/s
B. 20ms
C. 30m/s
D. 40m/s
SOLUTION
S = vt+ = 0 +
= 50 m
Ave velocity = 50/5 = 10 m/s
PROBLEM 33
• From a speed of 75kph, a car
decelerates at the rate 0f 500m/min2
along a straight path. How far in
meters will it travel in 45 s?
A. 790.293m
B. 791.293m
C. 796.875m
D. 793.238m
SOLUTION
V2 = v1 .at
V1 = 75 kph (75000/ 60) = 1250 m/min
V2 = v1.at = 1250 – 500(45)/ 60 = 875 m/min
V22 = v12 - 2aS
(875)2 = (1250)2 - 2(500)S
S = 796.875 m
PROBLEM 34
• An object experiences rectilinear
acceleration a(t) = 10-2t. how far
does it travel in 6 sec if its initial
velocity is 10m/s?
A. 182
B. 168
C. 174
D. 154
SOLUTION
A=
Dv / dt = 10. 2t
Dv = (10 – 2t )dt
=
V - 10 =10t- 2t2/2 = 10t- t2
V = 10t – t2 + 10 – ds/dt
Ds = vdt
= 10t – t2 + 10)dt
S=
= 5(36) – (6)3 / 3 + 10 (6) = 168 m
PROBLEM 35
• An object is accelerating to the right
along a straight path at 2m/s. the
object begins with a velocity of
10m/s to the left. How far does it
travel in 15 sec?
A. 125m
B. 130m
C. 140m
D. 100m
SOLUTION
V2 = v1 – at
0 = 10 – 2t
T = 5 sec
V22 = v12 - 2aS
0 = (10)2 - 2(2)S
S = 25 m
S = vt+
S = 0+
= 100 m
Total distance traveled = 125 m
PROBLEM 36
• A ball is dropped from a balloon at a
height of 195m. if the balloon is
rising 29.3m per sec , find the
highest point reached by the ball
and the time of flight.
A. 238.8m
B. 487.3m
C. 328.4m
D. 297.3m
SOLUTION
V22 = v12 - 2gh
0 = (29.3)2 - 2(9.81)h
H = 43.8m
H = 195 +43.8 = 238.8 m
PROBLEM 37
• An object falls a height of 92m and
strikes the ground with a speed of
19m/s . determine the height that
the object must fall in order to strike
with a speed of 24m/s.
A. 146.94m
B. 184.29m
C. 110.12m
D. 205.32m
SOLUTION
Vf = 19 , s = 92
Vf2 = Vi2 + 2aS
(19)2 = 0 + 2a(92)
A = 1.96 m /s2
When vf = 24
(24)2 = 0 - 2(1.96)S
S = 146.94 m
PROBLEM 38
• A ball drop freely from a balloon at a
height of 195m if the balloon is rising
29.3m/s. find the highest point
reached by the ball and the velocity
of the ball as it strikes the ground.
A. 43.76m, 68.44m/s
B. 22.46m, 71.66m/s
C. 36.24m, 69.24m/s
D. 12.80m, 31.20m/s
SOLUTION
V22 = v12 - 2gh
0= (29.3)2 - 2(9.81)3h
H = 43.76 m
H = 195 + 43.76 = 238.76 m
V32 = V22 + 2gh = 0 +2(9.81)(238.76)
V3 = 68.44 m/s
PROBLEM 39
• A projectile is fired from the top of a
cliff 92m high with a velocity of 430
m/s directed 45o to the horizontal.
Find the range on the horizontal
plane through the base of the cliff .
A. 18.940km
B. 23.408km
C. 15.273km
D. 20.365km
SOLUTION
Y = xtan 45 -92 =xtan 45 –
0.000053x2 – x – 92 = 0
X = 18959.48 m
PROBLEM 40
• A stone was thrown upward at an angle of
60o with the horizontal and a resultant
vertical speed of 100 m per sec . if the
gravity decelerates the speed at 9.8 m/s2,
what is the actual speed of the stone in m/s
, 10 sec later in the direction it was thrown?
A. 57.77m/s
B. 64.22m/s
C. 60.35m/s
D. 67.23m/s
SOLUTION
Voy = 100 m/ s
Tan 60 = Voy­/ V­ox­
V­ox­ = 100 / tan 60 = 57.74 m /s
After 10 s
V­y = V­oy + Aoy­­ = 100 +(-9.8)(10) = 2 m/s
V­x = V­ox­ = 57.74
V=
=
= 57.77 m/
PROBLEM 41
• A block passes a point 4m from the edge of
the table with a velocity of 5m/s . it slides
off the edge of a table which is 5 m high
and strikes the floor 3m from the edge of
the table. What was the coefficient of
friction between the block and the table?
A. 0.65
B. 1.04
C. 0.21
D. 0.11
SOLUTION
Y=
5=
T = 1.01 s
X = V2 t
3 = V2 (1.01)
V2 = 2.97 m/s
V22 = v12 - 2as
(2.92)2 = (5)2 - 2a(4)
A = 2.02 m/ s2
F = Wa / g
µN = wa / g = w (2.02) / 9.81
µ = 0.21
PROBLEM 42
• A ball is shot at a ground level at an
angle of 60 o with the horizontal with
an initial velocity of 10m/s . which of
the following most nearly gives the
maximum height attained by the ball?
A. 2.47m
B. 3.29m
C. 4.61m
D. 3.82m
SOLUTION
V22 = V12 - 2gh
0 = (10sin60)2 - 2(9.81)h
H= 3.82 m
PROBLEM 43
• A stone is thrown upward at an angle of
30o with the horizontal. It lands 60m
measured horizontally and 2m below
vertically from its point of release .
determine the initial velocity of the
stone in m/s.
A. 27.35
B. 28.35
C. 25.35
D. 26.35
SOLUTION
Y = xtan o -2 = 60tan 30 –
36.64 = V = 25.35 m/s
PROBLEM 44
• A projectile leaves a velocity of
50m/s at an angle of 30o with the
horizontal. Find the maximum
height that it could reach in meter.
A. 31.86
B. 41.26
C. 28.46
D. 51.26
SOLUTION
V22 = V12 - 2gh
0 = (50sin30)2 - 2(9.81)h
H = 31.86 m
PROBLEM 45
• A shot is fired at an angle of 45o
with the horizontal and a velocity
of 300 fps . calculate the range of
the projectile.
A. 3500 ft
B. 1200 ft
C. 4000 ft
D. 932 yds
SOLUTION
R = V2sin2o / g = (300)2 sin90 / 32.2 = 2795 ft =
2795 / 3 = 932 yds
PROBLEM 46
• A homogeneous sphere rolls down
an inclined plane making an angle of
30o with the horizontal. Determine
the minimum value of the coefficient
of friction which will prevent slipping.
A. 0.165
B. 0.362
C. 1.025
D. 0.525
SOLUTION
I = 2wr2 / 5g
A = rα
α = a/r
Fr = Iα = 2wr2(a) / 5gr
F = 2wa/5g
Wsin30 = f = wa /g = 2wa /5g + wa /g
0.5 = 7a /5g
A = (0.5)(5)(9.81)/7 = 3.504 m/s2
F = 2w(3.504)/ 5 (9.81) = 0.143 w
N = wcos30
F = µN
0.143w = µwcos30
µ = 0.165
PROBLEM 47
• Find the angle of banking for a
highway curve of 90 m radius designed
to accommodate cars traveling at 160
kph, if the coefficient of friction
between the tires and the road is 0.60.
A. 34.9 o
B. 44.6 o
C. 20.3 o
D. 47.2o
SOLUTION
Tan o = 0.60
O = 31o
V = 160 kph (1000/3600) = 44.44 m / s
Tan(o + a) = v2 / gr
Tan(o +31) = (44.44)2 / 9.81(90)
O +31 = 65.9
O = 34.9o
PROBLEM 48
• A cyclist moves at a constant speed
of 24m/s around a curve road having
a radius of 200 m. determine the
magnitude
of
the
cyclists
acceleration in m/s2.
A. 3.22
B. 1.49
C. 2.88
D. 4.12
SOLUTION
At = 0
An = v2 /r = (24)2 / 200 = 2.88 m / s2
A=
=
= 2. 88 m/s2 (towards the
center)
PROBLEM 49
• The angular speed of a rotating
flywheel a radius of 0.5m is 180/π
rpm. Compute the value of its
normal
acceleration
and
its
tangential speed.
A. 16m/s2 ; 2 m/s
B. 18m/s2 ; 3 m/s
C. 14m/s2 ; 1.5 m/s
D. 12m/s2 ; 1 m/s
SOLUTION
W = 180 2 / π 60
W = 6 rad /sec
An = rw2
An = 0.5(6)2
An = 18 m/s2
V = rw
V = 0.5(6)
V = 3m/s
PROBLEM 50
• An object is placed 3ft from the
center of a horizontally rotating
platform. The coefficient of friction is
0.30. the object will begin to slide off
when the platform speed is nearest to
A. 12rpm
B. 17rpm
C. 22rpm
D. 26rpm
SOLUTION
F =wv2 /gr
V =rw
0.30w = w (rw)2 / gr
W2 = 0.32(32.2) / 3
W = 1.79 rad /sec
W = 1.79(60) / 2π
W = 17.1 rpm
PROBLEM 51
• A tripod whose legs are each 4 m long
supports a load of 1000 kg. the feet of
the tripod are at the vertices of a
horizontal equilateral triangle whose
side is 3.5 m. determine the load on each
leg.
A. 386.19kg
B. 347.29kg
C. 214.69kg
D. 446.27kg
SOLUTION
Cos 30 = 1.75/ae
Ae = 2.02 m
Caso = 2.02/4
O = 59.67o
Ra sin59.67 + Rb sin59.67 + Rc59.67 = 1000
Ra = Rb= Rc
Ra (3sin59.67) = 1000
Ra = 386.19 kg
Rb = 386.19 kg
Rc = 386.19 kg
PROBLEM 52
• A point on the rim of a rotating flywheel
changes in speed from 1.5m/s to 9m/s
while it moves 60m. if the radius of the
wheel is 1m. compute the normal
acceleration at the instant when its
speed is 6m/s.
A. 36m/s2
B. 24m/s2
C. 18m/s2
D. 20m/s2
SOLUTION
V22 = V12 +2as
92 = 1.52 + 2a(60)
A= 0.656 m/s2 (tangential acceleration)
A = V2 / r = 62 / 10 = 36 m/s2 (normal
acceleration)
PROBLEM 53
• Starting from rest, an elevator weighing
9000N attains an upward velocity of 5m/s in
4 sec with uniform acceleration. Find the
apparent weight of a 600 N man standing
inside the elevator during its ascent and
calculate the tension in the supporting cable.
A. 10.823 N; 676.5N
B. 11.382 N; 700N
C. 9.254 N; 400 N
D. 12.483 N 725N
SOLUTION
V2= V1 + at
5 = 0 + a4
A = 1.25m/s2
R =600 + 600a/g = 600 + 600(1.25)/ 9.81
= 676.5 N
T = 9600 + 9600(1.25) / 9.81 = 10823 N
PROBLEM 54
• An elevator technician weighs 160 lbs in
an elevator which is at rest . when the
elevator moves up at a constant
acceleration , the technician weighs 200
lbs . how fast is the elevator
accelerating?
A. 8.05fps2
B. 7.06fps2
C. 6.06fps2
D. 5.06fps2
SOLUTION
160 +f = 200
F = 40lb = ma = w/g (a)
40 = 160 /32.2 (a)
A = 8.05 fps2
PROBLEM 55
• The tensile capacity of the rope is
1500N . if this rope is used to tow a
500kg car on a smooth level road, what
is the greatest acceleration that the car
can move without breaking the ropes?
A. 3
B. 4
C. 5
D. 6
SOLUTION
F = ma
1500 = 500a
A= 3 m/s2
PROBLEM 56
• A hauling truck weighing 14000 N is moving
up a slope of 20o. if the coefficient of
friction between the truck tires and the
pavement on a slope 20o is 0.20. what force
will be generated by the engine if the truck
is accelerating at the rate of 3m/s2?
A. 12.3kN
B. 11.7kN
C. 10.8kN
D. 13.4kN
SOLUTION
F1 = µN
F1 = 0.20 wcos o
F1 = 0.20(14000)sin20 = 2631 N
Wsin o = 14000sin20 = 4788 N
F = 4788 – 2631 = 14000/9.81 (3)
F = 11700 N
F = 11.7kN
PROBLEM 57
• A 1N bullet is fired with a velocity of
900m/s into an 89 N sand box mounted
on wheels. After the impact, the box
strikes a spring and is brought to rest in a
distance of 150mm. find the spring
constant.
A. 40.8N/mm
B. 45.2N/mm
C. 33.6N/mm
D. 52.7N/mm
SOLUTION
M1v1 +m2v2 = m1v1 +m2v
1(900) /g + 89(0) / g + 1v/g+ 89V/g
90v = 900
V = 10 m/s
Ks2 / 2 = w/2g (v22 – v12)
K(150)2 / 2(1000) = 90 / 2(9.81) (02 – 102)
K = 40.8 N/mm
PROBLEM 58
• A 5o kg object strikes the unstretched
spring attached to a vertical wall having a
spring constant of 20kN/m . find the
maximum deflection of the spring. The
velocity of the object before it strikes the
spring is 40m/s.
A. 1m
B. 2m
C. 3m
D. 4m
SOLUTION
1/2mv2 = 1/2kx2
½(50)(40)2 = ½ (20000)x2
X = 2m
PROBLEM 59
• A large coil spring with a spring
constant k = 120N/m is elongated
within its elastic range by 1m.
compute the stored energy of the
spring in Nm.
A. 60
B. 40
C. 50
D. 120
SOLUTION
F = kx – 120x
Energy =
=
= 60 N.m
20x2 /2
PROBLEM 60
• A truck weighing 1000kN is being pulled
up a 2% grade . the train resistant is 5
kN/m. the speed of the truck is increased
from 6m/s to 12m/s in a distance of
300m . find the maximum horsepower
developed by the locomotive.
A. 706 hp
B. 697 hp
C. 844 hp
D. 775 hp
SOLUTION
Pos work – neg work = change in ke
Sin o = tan o
300p – 5000(300) – 1000(1000)2/100 (300) = ½
(1000)2 /9.81 (144-36)
P = 43349N
Hp = pv / 746 = 43349(12) / 746 = 697 hp
PROBLEM 61
• A ball strikes the ground at an angle of
30o with the ground surface, the ball
then rebounds at a certain angle φ with
the ground surface. If the coefficient of
restitution is 0.80, find the value of the
angle.
A. 24.79o
B. 18.48o
C. 32.2o
D. 26.7o
SOLUTION
E= Vsin o / Vsin 30
V = V cos o / cos 30
E = vsin0cos30 / vcos0sin30 = tanocot30
0.80 = tan0cot30
O = 24.79
PROBLEM 62
• A man weighing 68kg jumps from a pier
with a horizontal velocity of 6m/s onto a
boat that is rest on the water. If the boat
weighs 100kg , what is the velocity of the
boat when the man comes to rest
relative to the boat?
A. 2.43m/s
B. 3.53m/s
C. 2.88m/s
D. 1.42m/s
SOLUTION
M1v1+m2v2= m1v1+m2v1
68(6) +100(0) = 68v +100v
V = 2.43 m/s
PROBLEM 63
• A ball is dropped from a height of
20m upon a stationary slab. If the
coefficient of restitution is 0.40,
how high will the ball rebound?
A. 3.2m
B. 4.6m
C. 5.2m
D. 8.0m
SOLUTION
E = v1/v2
0.40 =
H = 3.2 m
=
PROBLEM 64
• A 60 ton rail car moving at 1 mile/hr
instantaneously coupled to a
stationary 40 ton rail car. What is
the speed of the coupled cars?
A. 0.88mi/hr
B. 1.0mi/hr
C. 0.6mi/hr
D. 0.40mi/hr
SOLUTION
M1v1+m2v2= m1v1+m2v1
60(1) + 40(0) = (60 + 40) v
V = 0.60 mph
PROBLEM 65
• A 1500 N block is in the contact with a
level plane whose coefficient of kinetic
friction is 0.10. if the block is acted upon
a horizontal force of 250 n at what time
will elapse before the block reaches a
velocity of 14.5m/s starting from rest?
A. 22.17 s
B. 18.36 s
C. 21.12 s
D. 16.93 s
SOLUTION
F = µN = 0.10(1500) = 150N
= m(v2 –v1) = w/g (v2 –v1)
(250 -150)t = 1500/9.81(14.5 -0)
T =22.17 s
PROBLEM 66
• A 1600 N block is in the contact with a level
plane whose coefficient of kinetic friction is
0.20. if the block is acted upon a horizontal
force of 300 N initially when the block is at
rest and the force was then removed when
the velocity of the block reaches 16m/s, how
much longer will the block continue to move ?
A. 8.15s
B. 6.25s
C. 4.36s
D. 5.75s
SOLUTION
F = µN = 0.20(1600) = 320N
= m(v2 –v1) = w/g (v3 –v2)
(0 -320)t = 1600/9.81(0 -16)
T = 8.15 s
PROBLEM 67
• A 50 kg block initially at rest is acted
upon by a force P which varies. Knowing
that the coefficient of kinetic friction
between the block and the horizontal
surface is 0.20 . compute the velocity of
the block after 5 sec and after 8 sec.
A. 15.19m/s;16.804m/s
B. 13.23m/s;15.534m/s
C. 10.65m/s;17.705m/s
D. 17.46m/s;14.312m/s
SOLUTION
T =0
= mv
250(5) = 10(9.81)(5) – 50 (v -0)
V = 15.19 m/s
250(5) + 250(3) / 2 – 10(9.81)(8) = 50 (v-0)
V = 16.804 m/s
PROBLEM 68
• A wood block weighing 44.75 N rests on a
rough horizontal plane , the coefficient of
friction being 0.40. if a bullet weighing 0.25 N
is fired horizontally into the block with a
velocity of 600m/s, how far will the block be
displaced from its initial position? Assume
that the bullet remains inside the block.
A. 1.41m
B. 2.42m
C. 1.89m
D. 0.98m
SOLUTION
M1v1+m2v2= m1v1+m2v1
0.25(600) /g + 44.75(0) /g = .25v /g +44.75v /g
45v = 0.25(600)
V = 3.33 m/s
Pos work – neg work = change in ke
O = 0.40(45)x = 45/2(9.81) (02 -3.332)
X = 1.41 m
PROBLEM 69
• A ball is dropped from a height of
20 m upon a stationary slab. If the
coefficient of restitution is 0.40,
how high will the ball rebound?
A. 3.2m
B. 4.6m
C. 5.2m
D. 8.0m
SOLUTION
E = v1/v2
0.40 =
H = 3.2 m
=
PROBLEM 70
• When the angular velocity of a 4 m
diameter pulley is 3 rad/sec , the total
acceleration of a point on its rim is
30m/s2. Determine the angular
acceleration of the pulley at this instant
A. 12rad/sec2
B. 10rad/sec2
C. 11rad/sec2
D. 9rad/sec2
SOLUTION
A = 30 m/s2
W = 3rad /sec
R = 4m
A=
A2 = at2 -ai2
At = rα
302 = r2 α2 + r2 w4
302 = 42 α2 + 42 34
α = 12 rad / sec2
END……