Download Exam (Fall16) 1-5

PHY 131
Ch 1 to 5
A stone is dropped from a
cliff. The graph that best
represents its motion while
it falls is:
III
A large cannon is fired from ground level over level ground
at an angle of 30° above the horizontal. The muzzle speed
is 980 m/s. The projectile will travel what horizontal dist
before striking the ground?
y = ½ a t2 +
vo
t + yo
0 = ½-10t2 + si n30(980) t + 0
t = 98 sec
One object is thrown vertically upward with an initial
velocity of 100 m/s and another object with an initial
velocity of 10 m/s. The maximum height reached by the
first object will be _______ that of the other.
a = ∆v / ∆t
-10 = 0 – 100 / t
t = 10 sec
a = ∆v / ∆t
-10 = 0 – 10 / t
t = 1 sec
y = ½ a t2 + vo t
y = ½-10 t2 + 100t
y = 500 m
y = ½ a t2 + vo t
y = ½-10 t2 + 10t
y=5m
An object placed on an equal-arm balance requires 12 kg to
balance it. When placed on a spring scale, the scale reads
12 kg. Everything is now transported to the Moon where
free-fall acceleration is 1/6 that on Earth. The new
readings of the balance and spring scale (respectively) are:
-kx = mg (gmoon = 1.66 m/s2)
Ans: 12 kg, 2 kg
Two blocks weighing 250 N and 350 N
respectively, are connected by a string
that passes over a massless pulley as shown.
The tension in the string is:
FNet
= mT
a
350 – 250 = (25+35) a
a = 1.7 m/s2
Accel down
Accel UP
FT = 35kg (10-1.7) OR 25kg (10+1.7)
FT = 290 N
An object is thrown straight down with an initial speed of 4
m/s from a window which is 8 m above the ground. The time
it takes the object to reach the ground is:
y = ½ a t + vot + yo
0 = ½-10 t2 - 4t + 8
t = 0.93 sec
2
A Ferris wheel with a radius of 8.0 m makes 1 revolution
every 10 s. As he passed the top of the path he releases a
ball. How far from the point on the ground directly under
the release point does the ball land?
h = ½ a t2
vx = ∆x / ∆t
v = 2πr / T
2r = ½10 t2
5 = ∆x / 1.8
v = 2π8 / 10
tbottom = 1.8 sec
∆x = 9 m
v = 5 m/s
A ball is thrown horizontally from the top of a 20-m high
hill. It strikes the ground at an angle of 45°. With what
speed was it thrown?
y = ½ a t2
20 = ½-10 t2
t = 2 sec
a =
∆vy / ∆t
-10 = (si n45v – 0)/2
v = 28 m/s
Name:
vx
= ∆x /∆t
cos30 (980) = ∆x /98
∆x = 83000 m
A ferry boat is sailing at 12 km/h 30° W of N with respect
to a river that is flowing at 6.0 km/h E. As observed from
the shore, the ferry boat is sailing:
vx
vy
No x component
River
6
0
Due North
Ferry
-sin30 12
cos30 12
Sum
0
10.4
When a 40-N force, parallel and directed up a ramp inclined
by 30° above the horizontal, the acceleration of the crate
is 2.0 m/s2, up the incline. The mass of the crate is:
FNet
= mT a
40 – sin30 mg = m (2)
m = 5.7 kg
A 25-kg crate is pushed across a frictionless horizontal
floor with a force of 20 N, directed 20° below the
horizontal. The accel of the crate is:
FNet
= mT a
cos20 (20)= 25 a
a = 0.75 m/s2
A 5-kg block is suspended by a rope from
the ceiling of an elevator that accelerates
downward at 3.0 m/s2. The tension force
of the rope on the block is:
FT = m (g + a)
FT = 5 (10 + 3)
FT = 65 N, up
The position of an object as a function of time is given in
meters by x = (at +bt2) i + (ct) j. What is its velocity as a
function of time?
v = dx / dt
v = (a + 2bt) i + (c) j
An object is thrown vertically into the air. Which of the
following five graphs represents the velocity (v) of the
object as a function of the time (t)? (Up is positive)
III
A projectile is shot vertically upward with a given initial
velocity. It reaches a maximum height of 100 m. If on a
second shot the initial velocity is doubled, then the
projectile will reach a maximum height of:
y=½
a t2
100 = ½-10 t2
t = √20
a = ∆v / ∆t
-10 = 0 – 90 / t
t = 9 sec
a = ∆v
/ ∆t
-10 = 0 – vo / √20
vo = 45 m/s
y = ½ a t2 + vo t
y = ½-10 t2 + 90t
y = 400 m
When a 40-N force, parallel and directed up a ramp inclined
by 30° above the horizontal, the acceleration of the crate
is 2.0 m/s2, down the incline. The mass of the crate is:
FNet
= mT a
40 – sin30 mg = m (-2)
m = 13.8 kg
v 2 = v x2 + v y2
v 2 = (14-6)2 + 62
v = 10 mph
A boat is traveling upstream at 14 mph
with respect to a river that is flowing at
6 mph. A man runs directly across the
boat, from one side to the other, at 6
mph. The speed of the man with respect
to the ground is:
A girl jogs around a horizontal circle with a constant speed.
She travels one fourth of a revolution, a distance of 25 m
along the circumference of the circle, in 5.0 s. The
magnitude of her acceleration is:
¼(2πr) = 25m
a = v2 / r
r = 50/π
a = (25/5)2 / 15.9
r = 15.9 m
a = 1.6 m/s2
A dart is thrown horizontally at 20 m/s. It
hits 0.1 s later. What is the distance the
dart drops during its path?
y = ½ a t2
y = ½10 .1 2
y = 0.05 m
Two forces, one with a magnitude of 3 N and the other with
a magnitude of 5 N, are applied to an object. For which
orientation of the shown forces shown is the magnitude of
the acceleration of the object the least?
We definitely
know:
IV is
Greatest,
and
I is the Least
A baseball is hit straight up and is caught by
the catcher 2.0 s later, at the same height at
which it left the bat. The maximum height of
the ball during this interval is:
y=½at
y = ½10 1
y=5m
2
top
2
A crate rests on a horizontal surface and a woman pulls on
it with a 10-N force. Rank the situations shown below
according to the magnitude of the normal force exerted by
the surface on the crate, least to greatest.
3, 2, 1
A car drives in a straight line on a level
road with a constant acceleration of 3
m/s2. A ball is suspended by a string
from the ceiling of the car; the ball
does not swing, being at rest with
respect to the car. What angle does
the string make with the vertical?
tan θ = 3/10
θ = 17°
A 90-kg man stands in an elevator that
is moving up at a constant speed of 5.0
m/s. The force exerted by him on the
floor is about:
FW = m (g + a)
FW = 90 (10 + 0)
FW = 900N
A massless rope passes over a massless pulley suspended
from the ceiling. A 4-kg block is attached to
one end and a 5-kg block is attached to the
other end. The acceleration of the 5-kg
block is:
FNet
= mT
a
5g – 4g = (4+5) a
a = g/9 m/s2
An object is dropped from rest.
Which of the five following graphs
correctly represents its motion?
The positive direction is taken to
be upward. Take careful note of
the axes.
An object is thrown vertically upward with a certain initial
velocity in a world where the acceleration due to gravity is
19.6 m/s2. The height to which it rises is ____ that to
which the object would rise if thrown upward with the same
initial velocity on the Earth. half
a = ∆v / ∆t
-10 = 0 – 100 / t
t = 10 sec
y = ½ a t2 + vo t
y = ½-10 t2 + 100t
y = 500 m
a = ∆v / ∆t
-20 = 0 – 100 / t
t = 5 sec
y = ½ a t2 + vo t
y = ½-20 t2 + 100t
y = 250 m
A short 10-g string is used to pull a 50-g toy across a
frictionless horizontal surface. If a 0.030 N force is
applied horizontally to the free end of the string, the force
of the string on the toy, at the other end, is:
Newton’s 3 rd law: 0.030 N
A motor boat can travel at 10 km/h in still water. A river
flows at 5 km/h west. A boater wishes to cross from the
south bank to a point directly opposite on the north bank.
At what angle must the boat be headed?
tan θ = 5 / √75
vx
vy
θ = 30°
River
5
0
Boat
-5
102 = -52 + v y2
Sum
0
v y = √75
A stone is thrown vertically upward with an initial speed of
19.5 m/s. It will rise to a maximum height of:
A stone is tied to a 0.50-m string and whirled
at a constant speed of 4.0 m/s in a vertical
circle. The acceleration at the bottom of the
circle is:
a = v 2/r = 32m/s2, up
FT – mg = mv 2/r
FT = 5mg
a = ∆v / ∆t
-10 = 0 – 19.5 / t
t = 1.95 sec
A boy on the edge of a vertical cliff 20 m high throws a
stone horizontally outwards with a speed of 20 m/s. It
strikes the ground at what horizontal distance from the
foot of the cliff?
y=½
a t2
20 = ½10 t2
t = 2 sec
vx = ∆x / ∆t
20 = xo – 0 / 2
xo = 40 m
A constant force of 8.0 N is exerted for 4.0 s on a 16-kg
object initially at rest. The change in speed of this object
will be:
F= ma
a = ∆v /∆t
8 = 16 a
½ = vf – 0 / 4
a = ½ m/s2
vf = 2 m/s
y = ½ a t2 + vo t
y = ½-10 t2 + 19.5t
y = 19 m
A 100 gram pendulum bob is
held at an angle θ from the
vertical by a 2 N horizontal
force F. The tension in the
string is:
tan θ = 2/1
θ = 63°
FT2 = 22 + 12
FT = √5
Two forces are applied to a 5.0-kg crate;
one is 6.0 N to the north and the other is
8.0 N to the west. The magnitude of the
acceleration of the crate is:
A particle goes from
x = –2 m, y = 3 m, z = 1 m
x = 3 m, y = –1 m, z = 4 m.
Its displacement is:
to
a 2 = 6/5 2 + 8/52
a = 2 m/s2
(3 - -2), (-1 – 3), (4 - 1)
(5, -4, 3)
A 32-N force, parallel to the incline, is required to push a
certain crate at constant velocity up a frictionless incline
that is 30° above the horizontal. The mass of the crate is:
FNet
= mT a
32 – sin30 mg = m (0)
m = 6.4 kg
(45%) Below is a ramp inclined by 30.0 degrees, initially at rest. A
Force, F = 30.0 N, is applied by a rocket (unlimited fuel). The 2.00
block will travel another 3 meters up the ramp before going over the
edge. The coordinates of the edge are (0, 0), up  + and right  +.
(a) What is the normal
2%
2% 8%
force exerted by the
FN = cos30(20) + F sin30
ramp on the block?
FN = 32 Newtons
(b) If a landing pad is 1 meter below
the edge, how far will the block travel in the horizontal direction?
8%
FNet
3%
= mT a
2%
1%
Fcos30 – sin30 mg = m a
26
- 10
= 2 (a)
a = 8 m/s2
x = ½ a t2
2%
3 = ½ 8 t2
t = 0.866 sec
On Ramp
Free Fall with Rocket
a = ∆v / ∆t
8 = vf – 0 / .866
vf = 6.93 m/s 2%
vx = cos30(6.93)
vx = 6 m/s
1%
Fx = m ax
30 = 2 (ax )
ax = 15 m/s2
vy = si n30(6.93)
vy = 3.46 m/s
1%
5%
y = ½ a t2 + vo t + yo
1% 1% 1% 1%
0 = ½-10t2 + 3.46 t + 1
t = 0.91 sec
x = ½ a t2 + vo t + yo
1% 1% 1% 1%
x = ½ 15 .91 2 + 6(.91) + 0
x = 13.1 meters
F = 20 N
(a)
2% 2%
8%
FN = 20N + sin30 F
FN = 30 Newtons
FNet
6%
1%
Fcos30 = m a
17.3
= 2 (a)
a = 8.66 m/s2
(a)
2% 2%
8%
FN = 20cos30 + sin30 F
FN = 21.6 Newtons
On Ramp
(b)
2%
FNet
3%
8%
= mT a
1%
mg sin30 - Fcos30 = m a
3.07
= 2 (a)
2
a = 1.5 m/s
x = ½ a t2
3 = ½ 1.54 t2
t = 2.0 sec
2%
Free Fall
a = ∆v / ∆t
8.66 = vf – 0 / .832
vf = 7.21 m/s 2%
Fy
= m ax
7%
1% 1%
si n30F + mg = 2 (ay)
ay = -15 m/s2
x = ½ a t2
3 = ½ 8.66 t2
t = 0.832 sec
= mT a
F = 8.00 N
On Ramp
2%
Fx
= m ax
cos30 F = 2 (ax )
ax = 8.66 m/s2
5%
End of Ramp
y = ½ at2 + vot + yo
1%
1% 1%
0 = ½-15t2 + 0 + 1
t = 0.365 sec
Free Fall
a = ∆v / ∆t
1.5 = vf – 0 / 2.0
vf = 3.0 m/s 2%
Fx = m ax
-8 = 2 (ax )
ax = -4 m/s2
vx = cos30(3)
vx = 2.6 m/s
y = ½ a t2 + vo t + yo
1% 1% 1% 1%
0 = ½10t2 + 1.5 t - 1
t = 0.321 sec
1%
vy = si n30(3)
vy = 1.5 m/s
1%
5%
x = ½ a t2 + vo t + yo
1% 1% 1% 1%
x = ½ -4 .321 2 + 2.6(.321) + 0
x = 0.64 meters
x = ½ a t2 + vo t + yo
2% 1% 1% 1%
x = ½ 8.66 .3652 + 7.21(.365) + 0
x = 3.21 meters